Electricity and Magnetism Practice Questions — JAMB UTME — Page 10

Question 181Question →

Two parallel resistors of values $6.0\text{ }\Omega$ and $12.0\text{ }\Omega$ are connected to a DC power source having an open-circuit voltage of $18.0\text{ V}$ and an internal resistance of $1.0\text{ }\Omega$ . What is the terminal potential difference supplied to the circuit?

14.4\text{ V}

18.0\text{ V}

3.6\text{ V}

17.0\text{ V}

Show answer & explanation

Answer: $14.4\text{ V}$

Answer

The terminal potential difference supplied to the circuit is

14.4\text{ V}

.

The option stating

14.4\text{ V}

is correct because the parallel combination of

6.0\text{ }\Omega

and

12.0\text{ }\Omega

gives an external equivalent resistance of

4.0\text{ }\Omega

. With an internal resistance of

1.0\text{ }\Omega

, the total circuit resistance is

5.0\text{ }\Omega

, resulting in a total current of

3.6\text{ A}

. The terminal voltage across the load is

3.6\text{ A} \times 4.0\text{ }\Omega = 14.4\text{ V}

.

Step-by-Step Solution

1

Calculate the equivalent resistance (

R_p

) of the parallel resistors.

R_p = \frac{6.0 \times 12.0}{6.0 + 12.0} = \frac{72.0}{18.0} = 4.0\text{ }\Omega

Resistors connected in parallel combine according to

\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}

.

2

Determine the total resistance (

R_{\text{total}}

) of the entire circuit, including internal resistance.

R_{\text{total}} = R_p + r = 4.0\text{ }\Omega + 1.0\text{ }\Omega = 5.0\text{ }\Omega

The internal resistance of the power source is in series with the external parallel load.

3

Calculate the total current (

I

) drawn from the power source.

I = \frac{E}{R_{\text{total}}} = \frac{18.0\text{ V}}{5.0\text{ }\Omega} = 3.6\text{ A}

Ohm's law for a complete circuit relates e.m.f., total resistance, and total current.

4

Calculate the terminal potential difference (

V

).

V = I \cdot R_p = 3.6\text{ A} \times 4.0\text{ }\Omega = 14.4\text{ V}

(or

V = E - I r = 18.0\text{ V} - 3.6\text{ V} = 14.4\text{ V}

)

Terminal voltage is the voltage across the external load or the e.m.f. minus the lost volts inside the cell.

Key Concept

Terminal Potential Difference and Internal Resistance

Question 182Question →

A galvanometer with an internal resistance of $100\ \Omega$ gives a full-scale deflection for a current of $1.0\text{ mA}$ . It is converted into an ammeter capable of measuring up to $1.0\text{ A}$ by connecting a suitable shunt resistor across it. This converted ammeter is then connected in series with a load resistor of $4.40\ \Omega$ and a cell of electromotive force $3.0\text{ V}$ having an internal resistance of $0.50\ \Omega$ . What is the current reading indicated by the ammeter?

0.60\text{ A}

0.67\text{ A}

0.68\text{ A}

0.03\text{ A}

Show answer & explanation

Answer: $0.60\text{ A}$

Answer

The ammeter will indicate a current of

0.60\text{ A}

.

To find the ammeter reading, first calculate the effective ammeter resistance

R_A = \frac{I_g G}{I} = \frac{0.001 \times 100}{1.0} = 0.10\ \Omega

. The total circuit resistance is

R_{\text{total}} = R_A + R + r = 0.10 + 4.40 + 0.50 = 5.00\ \Omega

. Applying Ohm's law

I = \frac{E}{R_{\text{total}}} = \frac{3.0}{5.00} = 0.60\text{ A}

.

Step-by-Step Solution

1

Calculate the equivalent resistance of the converted ammeter (

R_A

).

R_A = \frac{I_g \times G}{I_{\text{max}}} = \frac{0.001\text{ A} \times 100\ \Omega}{1.0\text{ A}} = 0.10\ \Omega

The shunt resistor is connected in parallel with the galvanometer so that the potential difference across both components is equal.

2

Determine the total resistance (

R_{\text{total}}

) of the complete series circuit.

R_{\text{total}} = R_A + R + r = 0.10\ \Omega + 4.40\ \Omega + 0.50\ \Omega = 5.00\ \Omega

The ammeter, load resistor, and internal resistance of the cell are all connected in series.

3

Calculate the circuit current (

I

) using Ohm's law for a complete circuit.

I = \frac{E}{R_{\text{total}}} = \frac{3.0\text{ V}}{5.00\ \Omega} = 0.60\text{ A}

The total current flowing through the circuit is the electromotive force divided by the total resistance.

Key Concept

Galvanometer Conversion to Ammeter and Circuit Current Calculation

Estimated Time:2m 0s

Question 183Question →

A $4.0\text{ }\mu\text{F}$ capacitor and a $6.0\text{ }\mu\text{F}$ capacitor are connected in parallel. This parallel combination is connected in series with a $10.0\text{ }\mu\text{F}$ capacitor across a $60.0\text{ V}$ d.c. power supply. What is the total energy stored in the combination?

9.0 \times 10^{-3}\text{ J}

3.6 \times 10^{-2}\text{ J}

3.48 \times 10^{-3}\text{ J}

2.23 \times 10^{-2}\text{ J}

Show answer & explanation

Answer: $9.0 \times 10^{-3}\text{ J}$

Answer

The total energy stored in the combination is

9.0 \times 10^{-3}\text{ J}

.

First, find the equivalent capacitance of the parallel branch by adding

4.0\text{ }\mu\text{F}

and

6.0\text{ }\mu\text{F}

to get

10.0\text{ }\mu\text{F}

. Next, combine this result in series with the

10.0\text{ }\mu\text{F}

capacitor to obtain a total circuit capacitance of

5.0\text{ }\mu\text{F}

(or

5.0 \times 10^{-6}\text{ F}

). Substituting

C_{eq} = 5.0 \times 10^{-6}\text{ F}

and

V = 60.0\text{ V}

into

E = \frac{1}{2} C_{eq} V^2

gives

9.0 \times 10^{-3}\text{ J}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the parallel branch.

C_p = C_1 + C_2 = 4.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 10.0\text{ }\mu\text{F}

Capacitors in parallel add directly.

2

Calculate the total equivalent capacitance of the entire circuit.

C_{eq} = \frac{C_p \times C_3}{C_p + C_3} = \frac{10.0\text{ }\mu\text{F} \times 10.0\text{ }\mu\text{F}}{10.0\text{ }\mu\text{F} + 10.0\text{ }\mu\text{F}} = 5.0\text{ }\mu\text{F} = 5.0 \times 10^{-6}\text{ F}

The parallel combination

C_p

is connected in series with

C_3

.

3

Calculate the total energy stored using

E = \frac{1}{2} C_{eq} V^2

.

E = \frac{1}{2} (5.0 \times 10^{-6}\text{ F}) (60.0\text{ V})^2 = 9.0 \times 10^{-3}\text{ J}

The energy stored in a capacitive circuit depends on the net capacitance and applied potential difference.

Key Concept

Equivalent capacitance of mixed networks and stored energy in capacitors

Question 184Question →

A driver cell of electromotive force $E_0 = 4.0\text{ V}$ and internal resistance $r = 1.0\ \Omega$ is connected across a uniform potentiometer wire of length $100\text{ cm}$ and total resistance $R_w = 9.0\ \Omega$ . A test cell of unknown electromotive force $E_x$ and internal resistance $r_x = 0.5\ \Omega$ is connected in series with a sensitive galvanometer in the secondary circuit. When the jockey touches the wire at a distance of $60\text{ cm}$ from the high-potential end, the galvanometer indicates zero deflection. What is the electromotive force $E_x$ of the test cell?

2.16\text{ V}

2.40\text{ V}

1.96\text{ V}

3.60\text{ V}

Show answer & explanation

Answer: $2.16\text{ V}$

Answer

The electromotive force of the test cell is

2.16\text{ V}

.

The correct answer is obtained by first including the driver cell's internal resistance to find the true driver current of 0.40 A. The total voltage across the 100 cm wire is therefore 3.60 V, yielding a potential gradient of 0.036 V/cm. Multiplying by the 60 cm balance length gives an EMF of 2.16 V for the test cell.

Step-by-Step Solution

1

Calculate total resistance in the primary driver circuit.

R_{\text{total}} = R_w + r = 9.0\ \Omega + 1.0\ \Omega = 10.0\ \Omega

The driver cell's internal resistance is in series with the potentiometer wire resistance.

2

Calculate the current flowing through the driver circuit.

I = \frac{E_0}{R_{\text{total}}} = \frac{4.0\text{ V}}{10.0\ \Omega} = 0.40\text{ A}

Ohm's law applied to the complete primary circuit gives the steady driver current.

3

Determine the potential drop across the entire 100 cm potentiometer wire.

V_w = I \times R_w = 0.40\text{ A} \times 9.0\ \Omega = 3.60\text{ V}

The potential drop across the wire depends on driver current and wire resistance.

4

Calculate potential gradient and solve for the unknown EMF

E_x

at the 60 cm balance point.

k = \frac{V_w}{L_{\text{total}}} = \frac{3.60\text{ V}}{100\text{ cm}} = 0.036\text{ V/cm}

, so

E_x = k \times L = 0.036\text{ V/cm} \times 60\text{ cm} = 2.16\text{ V}

At zero galvanometer deflection, no current flows through the test cell, so its terminal voltage equals its EMF

E_x

.

Key Concept

Potentiometer balance condition and primary circuit internal resistance considerations

Question 185Question →

A potentiometer wire of length $100\text{ cm}$ has a resistance of $5.0\ \Omega$ . It is connected in series with a driver cell of e.m.f. $3.0\text{ V}$ (having negligible internal resistance) and a protective series resistor $R_s$ . A balance point is obtained at $60.0\text{ cm}$ along the wire for a cell of e.m.f. $1.2\text{ V}$ . What is the value of the series resistor $R_s$ in ohms?

Show answer & explanation

Answer: 2.5

Answer

The resistance of the series resistor

R_s

is

2.5\ \Omega

.

At balance, the potential drop across the

60.0\text{ cm}

portion of the potentiometer wire equals the test cell e.m.f. (

1.2\text{ V}

). Since the

100\text{ cm}

wire has a total resistance of

5.0\ \Omega

, the

60.0\text{ cm}

section has a resistance of

3.0\ \Omega

. This requires a main circuit current of

I = \frac{1.2\text{ V}}{3.0\ \Omega} = 0.4\text{ A}

. The total driver circuit resistance is

R_{total} = \frac{3.0\text{ V}}{0.4\text{ A}} = 7.5\ \Omega

. Subtracting the wire's

5.0\ \Omega

resistance gives

R_s = 2.5\ \Omega

.

Step-by-Step Solution

1

Calculate the resistance of the balanced portion of the potentiometer wire.

R_{balance} = 5.0\ \Omega \times \frac{60.0\text{ cm}}{100\text{ cm}} = 3.0\ \Omega

The resistance of a uniform wire is directly proportional to its length.

2

Determine the current in the main potentiometer circuit.

I = \frac{1.2\text{ V}}{3.0\ \Omega} = 0.4\text{ A}

At the balance point, no current flows through the galvanometer branch, so the potential difference across the balance length equals the e.m.f. of the test cell.

3

Compute the total resistance of the driver circuit.

R_{total} = \frac{3.0\text{ V}}{0.4\text{ A}} = 7.5\ \Omega

According to Ohm's law, total resistance equals the total e.m.f. divided by the circuit current.

4

Calculate the value of the series resistor

R_s

.

R_s = 7.5\ \Omega - 5.0\ \Omega = 2.5\ \Omega

The total resistance is the sum of the potentiometer wire resistance and the series resistance.

Key Concept

Potentiometer balance condition and circuit analysis

Estimated Time:1m 30s

Question 186Question →

An electric lamp rated $60\,\text{W}$ operates normally when connected to a $240\,\text{V}$ mains supply. What is the electric current drawn by the lamp?

0.25\,\text{A}

4.0\,\text{A}

2.5\,\text{A}

14\,400\,\text{A}

Show answer & explanation

Answer: $0.25\,\text{A}$

Answer

The electric current drawn by the lamp is

0.25\,\text{A}

.

Electric power is defined by the formula

P = IV

, where

P

is power in watts,

I

is current in amperes, and

V

is potential difference in volts. Rearranging to solve for current yields

I = \frac{P}{V}

. Substituting

60\,\text{W}

for power and

240\,\text{V}

for voltage gives

I = \frac{60}{240} = 0.25\,\text{A}

.

Step-by-Step Solution

1

Identify given parameters and state the electric power formula.

Power

P = 60\,\text{W}

and voltage

V = 240\,\text{V}

. Power is related to current and voltage by

P = IV

.

Electric power dissipated by a component is the product of current and voltage across it.

2

Rearrange the formula to solve for current

I

.

I = \frac{P}{V}

Making current

I

the subject of the equation.

3

Substitute the values into the formula and solve.

I = \frac{60\,\text{W}}{240\,\text{V}} = 0.25\,\text{A}

Dividing

60

by

240

gives

0.25

amperes.

Key Concept

Relationship between Electrical Power, Voltage, and Current

Question 187Question →

A uniform copper wire with a cross-sectional area of $2.0 \times 10^{-6}\,\text{m}^2$ carries a steady electric current of $4.0\,\text{A}$ . Given that the free electron density of copper is $2.5 \times 10^{28}\,\text{m}^{-3}$ and the elementary charge is $1.6 \times 10^{-19}\,\text{C}$ , calculate the drift velocity of the electrons in the wire in $\text{m/s}$ .

Show answer & explanation

Answer: 0.0005

Answer

The drift velocity of the electrons in the wire is

0.0005\,\text{m/s}

(or

5.0 \times 10^{-4}\,\text{m/s}

).

Applying the formula

I = n A e v_d

and solving for drift velocity yields

v_d = \frac{I}{n A e} = \frac{4.0}{(2.5 \times 10^{28})(2.0 \times 10^{-6})(1.6 \times 10^{-19})} = \frac{4.0}{8000} = 0.0005\,\text{m/s}

.

Step-by-Step Solution

1

State the formula relating electric current to microscopic drift velocity.

I = n A e v_d

Electric current is the net charge flowing per unit time across a conductor's cross-section.

2

Isolate the drift velocity

v_d

as the target variable.

v_d = \frac{I}{n A e}

Algebraic rearrangement places all known quantities on the right-hand side.

3

Substitute the numerical values into the equation.

v_d = \frac{4.0}{(2.5 \times 10^{28})(2.0 \times 10^{-6})(1.6 \times 10^{-19})}

Inserting

I = 4.0\,\text{A}

,

n = 2.5 \times 10^{28}\,\text{m}^{-3}

,

A = 2.0 \times 10^{-6}\,\text{m}^2

, and

e = 1.6 \times 10^{-19}\,\text{C}

.

4

Evaluate the arithmetic computation.

v_d = 0.0005\,\text{m/s}

Dividing

4.0

by

8000

yields

0.0005\,\text{m/s}

.

Key Concept

Relationship between electric current, charge carrier density, and drift velocity

Question 188Question →

An electric lamp of resistance $10\,\Omega$ is connected to a cell of electromotive force (e.m.f.) $12\,\text{V}$ and internal resistance $2\,\Omega$ . What is the electrical power dissipated as heat inside the cell?

2.0\,\text{W}

2.88\,\text{W}

10.0\,\text{W}

12.0\,\text{W}

Show answer & explanation

Answer: $2.0\,\text{W}$

Answer

The electrical power dissipated inside the cell is

2.0\,\text{W}

.

The total opposition to current in the circuit includes both the external lamp resistance and the internal resistance of the cell (

10\,\Omega + 2\,\Omega = 12\,\Omega

). This yields a circuit current of

1.0\,\text{A}

. Using Joule's law of heating (

P = I^2 r

), the power lost specifically inside the cell is

(1.0)^2 \times 2 = 2.0\,\text{W}

.

Step-by-Step Solution

1

Calculate the total resistance of the circuit.

R_{\text{total}} = R + r = 10\,\Omega + 2\,\Omega = 12\,\Omega

The internal resistance of the cell is in series with the external load resistance.

2

Determine the current flowing through the circuit.

I = \frac{E}{R_{\text{total}}} = \frac{12\,\text{V}}{12\,\Omega} = 1.0\,\text{A}

By Ohm's law applied to a complete circuit, current equals total e.m.f. divided by total resistance.

3

Calculate the power dissipated as heat in the internal resistance.

P_{\text{internal}} = I^2 r = (1.0\,\text{A})^2 \times 2\,\Omega = 2.0\,\text{W}

Joule heating power in a resistor is given by

P = I^2 r

.

Key Concept

Electrical Power Dissipation and Internal Resistance

Estimated Time:1m 30s

Question 189Question →

Match each electrical quantity or unit in List I with its corresponding formula or definition in List II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electrical Power (

P

) matches Product of potential difference and current (

V \times I

); Electrical Energy (

E

) matches Product of electrical power and time (

P \times t

); SI unit of Electric Power matches Watt (

\text{W}

); Commercial unit of Electrical Energy matches Kilowatt-hour (

\text{kWh}

).

Electrical power is defined by the product of potential difference and current (

P = VI

) and measured in Watts (

\text{W}

). Electrical energy is power multiplied by time (

E = Pt

) and commercially measured in Kilowatt-hours (

\text{kWh}

).

Step-by-Step Solution

1

Identify the formula for electrical power.

Electrical power

P = VI

.

Power represents the rate of electrical energy dissipation across a potential difference.

2

Identify the formula for electrical energy.

Electrical energy

E = P \times t

.

Energy is the work done over a period of time.

3

Determine the standard SI unit for power.

The SI unit of power is the Watt (

\text{W}

).

One Watt is defined as one Joule per second.

4

Determine the commercial unit for electrical energy.

The commercial unit of energy is the Kilowatt-hour (

\text{kWh}

).

1 kWh is equal to

3.6 \times 10^6\,\text{J}

, which is convenient for commercial electricity billing.

Key Concept

Electrical Energy and Power Definitions and Units

Question 190Question →

A galvanometer has an internal resistance of $50\ \Omega$ and gives a full-scale deflection for a current of $2.0\text{ mA}$ . What multiplier resistance must be connected in series with the galvanometer to convert it into a voltmeter capable of measuring potential differences up to $10.0\text{ V}$ ?

4950\ \Omega

5000\ \Omega

5050\ \Omega

450\ \Omega

Show answer & explanation

Answer: $4950\ \Omega$

Answer

The required multiplier resistance is

4950\ \Omega

.

To convert a galvanometer to a voltmeter, a high-resistance multiplier

R_m

is connected in series. The maximum voltage

V

measured by the voltmeter is given by

V = I_g(G + R_m)

. Substituting

V = 10.0\text{ V}

,

I_g = 0.002\text{ A}

, and

G = 50\ \Omega

, we find

R_m = \frac{10.0}{0.002} - 50 = 4950\ \Omega

.

Step-by-Step Solution

1

Convert given values to standard SI units.

Galvanometer resistance

G = 50\ \Omega

, full-scale current

I_g = 2.0\text{ mA} = 0.002\text{ A}

, maximum voltage

V = 10.0\text{ V}

.

Electric current must be expressed in amperes (A) for standard circuit calculations.

2

Apply the series multiplier formula for a voltmeter.

V = I_g(G + R_m) \implies 10.0 = 0.002 \times (50 + R_m)

.

Converting a galvanometer to a voltmeter requires connecting a high resistance

R_m

in series so that the total voltage drop equals

V

.

3

Solve for the multiplier resistance

R_m

.

50 + R_m = \frac{10.0}{0.002} = 5000 \implies R_m = 5000 - 50 = 4950\ \Omega

.

Subtracting the internal resistance

G

from total resistance gives the value of the multiplier resistor alone.

Key Concept

Galvanometer Conversion to Voltmeter

Estimated Time:1m 30s

Question 191Question →

Two capacitors of capacitances $12.0\text{ }\mu\text{F}$ and $6.0\text{ }\mu\text{F}$ are connected in parallel. This combination is then connected in series with a $9.0\text{ }\mu\text{F}$ capacitor across a $45.0\text{ V}$ d.c. power supply. What is the total charge supplied by the power source?

270\text{ }\mu\text{C}

585\text{ }\mu\text{C}

1215\text{ }\mu\text{C}

125\text{ }\mu\text{C}

Show answer & explanation

Answer: $270\text{ }\mu\text{C}$

Answer

The total charge supplied by the power source is

270\text{ }\mu\text{C}

.

Combining the parallel capacitors yields

18.0\text{ }\mu\text{F}

. Connecting this combination in series with the

9.0\text{ }\mu\text{F}

capacitor gives an equivalent total capacitance of

6.0\text{ }\mu\text{F}

. Multiplying total capacitance by the supply voltage of

45.0\text{ V}

gives

270\text{ }\mu\text{C}

.

Step-by-Step Solution

1

Calculate equivalent capacitance of the parallel branch

C_{\text{p}} = C_1 + C_2 = 12.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 18.0\text{ }\mu\text{F}

Capacitors in parallel add directly.

2

Calculate total equivalent capacitance of the network

C_{\text{eq}} = \frac{C_{\text{p}} \times C_3}{C_{\text{p}} + C_3} = \frac{18.0 \times 9.0}{18.0 + 9.0} = 6.0\text{ }\mu\text{F}

The parallel combination

C_{\text{p}}

is in series with

C_3

, using the reciprocal formula for series capacitors.

3

Calculate total charge supplied

Q = C_{\text{eq}} V = 6.0 \times 10^{-6}\text{ F} \times 45.0\text{ V} = 270\text{ }\mu\text{C}

Total charge from the power source depends on total equivalent capacitance and total supply voltage.

Key Concept

Equivalent capacitance of mixed series-parallel networks

Estimated Time:1m 30s

Question 192Question →

A cell of electromotive force $E$ and internal resistance $r$ supplies a current of $1.2\text{ A}$ when connected across a $4.0\ \Omega$ resistor. When this resistor is replaced with a $10.0\ \Omega$ resistor, the current decreases to $0.6\text{ A}$ . Calculate the internal resistance $r$ of the cell in ohms ( $\Omega$ ).

Show answer & explanation

Answer: 2

Answer

The internal resistance of the cell is

2.0\ \Omega

.

For a complete circuit, the electromotive force

E

is related to load resistance

R

, internal resistance

r

, and current

I

by

E = I(R + r)

. Setting up equations for both load conditions gives

E = 1.2(4.0 + r)

and

E = 0.6(10.0 + r)

. Equating these expressions yields

4.8 + 1.2r = 6.0 + 0.6r

, which simplifies to

0.6r = 1.2

, giving an internal resistance of

r = 2.0\ \Omega

.

Step-by-Step Solution

1

Formulate the circuit equation for the first load resistance

E = 1.2 \times (4.0 + r) = 4.8 + 1.2r

Using Ohm's law for a complete circuit, the electromotive force equals current times total circuit resistance:

E = I(R_1 + r)

.

2

Formulate the circuit equation for the second load resistance

E = 0.6 \times (10.0 + r) = 6.0 + 0.6r

The cell maintains the same internal electromotive force

E

and internal resistance

r

with the new load resistor

R_2

.

3

Equate the expressions for

E

and solve for

r

4.8 + 1.2r = 6.0 + 0.6r \implies 0.6r = 1.2 \implies r = 2.0\ \Omega

Since

E

is constant for the cell, setting the two right-hand sides equal yields a single linear equation for the unknown internal resistance

r

.

Key Concept

Electromotive Force and Internal Resistance

Question 193Question →

Match each electrical scenario or description in List I with its corresponding mathematical expression or unit in List II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: Energy dissipated in duration

t

matches

I^2 R t

; Power rating under voltage

V

matches

\frac{V^2}{R}

; Commercial unit equal to

3.6 \times 10^6\,\text{J}

matches 1 kWh; Rate of heat production matches

I V

.

Each description in List I corresponds directly to its standard physical formula or commercial unit in List II based on the definitions of electrical power (

P = I V = \frac{V^2}{R}

) and electrical energy (

E = I^2 R t = P t

).

Step-by-Step Solution

1

Identify Joule's law of heating for energy dissipation over time.

Energy

E = I^2 R t

, matching the expression

I^2 R t

.

Joule's heating effect states that electrical energy transformed into heat energy is directly proportional to

I^2

,

R

, and

t

.

2

Relate power, voltage, and resistance.

Power

P = \frac{V^2}{R}

, matching the expression

\frac{V^2}{R}

.

Substituting Ohm's law

I = \frac{V}{R}

into power formula

P = I V

yields

P = \frac{V^2}{R}

.

3

Convert kilowatt-hours to joules for commercial unit matching.

1\,\text{kWh} = 1000\,\text{W} \times 3600\,\text{s} = 3.6 \times 10^6\,\text{J}

, matching 1 kWh.

Kilowatt-hour is the standard energy unit used by electric utility companies.

4

Determine the general definition of electrical power as the rate of energy transfer.

Power

P = I V

, matching the expression

I V

.

Electric power is defined as the product of current

I

and potential difference

V

.

Key Concept

Formulas and units for electrical energy, electric power, Joule's law of heating, and kilowatt-hours.

Estimated Time:1m 30s

Question 194Question →

A cell with an electromotive force (e.m.f.) of $15.0\text{ V}$ and an internal resistance of $2.0\ \Omega$ is connected across a parallel combination of two resistors with resistances of $6.0\ \Omega$ and $12.0\ \Omega$ . What is the potential difference across the parallel resistor network?

10.0\text{ V}

15.0\text{ V}

5.0\text{ V}

13.5\text{ V}

Show answer & explanation

Answer: $10.0\text{ V}$

Answer

The potential difference across the parallel resistor network is

10.0\text{ V}

.

The parallel combination of

6.0\ \Omega

and

12.0\ \Omega

has an effective resistance of

4.0\ \Omega

. Adding the internal resistance of

2.0\ \Omega

gives a total circuit resistance of

6.0\ \Omega

. The circuit current is

I = \frac{15.0\text{ V}}{6.0\ \Omega} = 2.5\text{ A}

. The potential difference across the parallel load is therefore

V = 2.5\text{ A} \times 4.0\ \Omega = 10.0\text{ V}

.

Step-by-Step Solution

1

Calculate the equivalent resistance of the two parallel resistors (

R_1 = 6.0\ \Omega

and

R_2 = 12.0\ \Omega

).

R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{6.0 \times 12.0}{6.0 + 12.0} = \frac{72.0}{18.0} = 4.0\ \Omega

Resistors connected in parallel combine according to the reciprocal formula.

2

Determine the total resistance of the entire circuit including the internal resistance (

r = 2.0\ \Omega

).

R_{\text{total}} = R_p + r = 4.0\ \Omega + 2.0\ \Omega = 6.0\ \Omega

The cell's internal resistance is in series with the external equivalent load resistance.

3

Calculate the total current supplied by the cell using Ohm's law for a complete circuit.

I = \frac{E}{R_{\text{total}}} = \frac{15.0\text{ V}}{6.0\ \Omega} = 2.5\text{ A}

The current depends on the total e.m.f. divided by the total circuit resistance.

4

Calculate the potential difference across the parallel combination (terminal potential difference).

V = I R_p = 2.5\text{ A} \times 4.0\ \Omega = 10.0\text{ V}

The potential difference across the parallel network is the product of the total current flowing into the combination and its equivalent resistance.

Key Concept

Terminal Potential Difference and Internal Resistance

Estimated Time:1m 30s

Question 195Question →

A voltmeter having an internal resistance of $900\ \Omega$ is connected across the terminals of a cell with an electromotive force (e.m.f.) of $1.50\text{ V}$ and an internal resistance of $100\ \Omega$ . What is the reading on the voltmeter?

1.35\text{ V}

1.50\text{ V}

0.15\text{ V}

0.17\text{ V}

Show answer & explanation

Answer: $1.35\text{ V}$

Answer

The reading on the voltmeter is

1.35\text{ V}

.

When connected across the cell, the voltmeter's resistance forms a series circuit with the cell's internal resistance. The total resistance is

900\ \Omega + 100\ \Omega = 1000\ \Omega

. The current drawn from the cell is

I = \frac{1.50\text{ V}}{1000\ \Omega} = 0.0015\text{ A}

. The voltage indicated by the meter is the potential difference across its terminals,

V = 0.0015\text{ A} \times 900\ \Omega = 1.35\text{ V}

(or

E - Ir = 1.50 - 0.15 = 1.35\text{ V}

).

Step-by-Step Solution

1

Calculate the total resistance of the circuit.

R_{\text{total}} = R_v + r = 900\ \Omega + 100\ \Omega = 1000\ \Omega

The voltmeter resistance and the internal resistance of the cell are connected in series.

2

Calculate the current flowing in the circuit using Ohm's law.

I = \frac{E}{R_{\text{total}}} = \frac{1.50\text{ V}}{1000\ \Omega} = 0.0015\text{ A}

The electromotive force drives current through the total circuit resistance.

3

Calculate the potential difference across the voltmeter (terminal potential difference).

V = I \times R_v = 0.0015\text{ A} \times 900\ \Omega = 1.35\text{ V}

The voltmeter measures the potential drop across its own internal resistance.

Key Concept

Terminal Potential Difference and Voltmeter Loading Effect

Estimated Time:1m 30s

Question 196Question →

A cell with an electromotive force (e.m.f.) of $12.0\text{ V}$ and an internal resistance of $1.5\ \Omega$ is connected in series to an external resistor of $8.5\ \Omega$ . What is the terminal potential difference across the cell?

Show answer & explanation

Answer: 10.2

Answer

The terminal potential difference across the cell is

10.2\text{ V}

.

The terminal potential difference

V

across a real cell supplying current is less than its electromotive force

E

due to the internal voltage drop

Ir

. By first determining the circuit current

I = \frac{E}{R + r} = \frac{12.0}{8.5 + 1.5} = 1.2\text{ A}

, the terminal potential difference is calculated as

V = I \times R = 1.2 \times 8.5 = 10.2\text{ V}

(or equivalently

V = E - I r = 12.0 - (1.2 \times 1.5) = 10.2\text{ V}

).

Step-by-Step Solution

1

Calculate the total resistance of the circuit

R_{\text{total}} = 10.0\ \Omega

The external load resistor and the cell's internal resistance are in series.

2

Determine the total current drawn from the cell

I = 1.2\text{ A}

Using Ohm's law applied to the entire circuit,

I = \frac{E}{R + r}

.

3

Calculate the potential drop across the external load resistor

V = 10.2\text{ V}

Terminal voltage equals potential difference across external load (

V = I R

) or

V = E - I r

.

Key Concept

Terminal Potential Difference and Internal Resistance

Question 197Question →

An electric immersion heater rated at $1.5\,\text{kW}$ is operated on a $240\,\text{V}$ mains supply for $14\,\text{minutes}$ . Calculate the total electrical energy consumed by the heater during this period, in megajoules ( $\text{MJ}$ ).

Show answer & explanation

Answer: 1.26

Answer

The total electrical energy consumed by the heater is

1.26\,\text{MJ}

.

Electrical energy consumed is obtained by multiplying electrical power by time (

E = P \times t

). Converting power to watts (

1.5\,\text{kW} = 1500\,\text{W}

) and time to seconds (

14\,\text{min} = 840\,\text{s}

) yields

E = 1500 \times 840 = 1,260,000\,\text{J} = 1.26\,\text{MJ}

.

Step-by-Step Solution

1

Convert power rating from kilowatts to watts

P = 1.5\,\text{kW} = 1500\,\text{W}

The standard SI unit of power for energy calculation in Joules is Watts.

2

Convert time duration from minutes to seconds

t = 14\,\text{minutes} \times 60\,\text{seconds/minute} = 840\,\text{seconds}

The standard SI unit of time in Joule calculations is seconds.

3

Calculate energy consumed in Joules using

E = P \times t

E = 1500\,\text{W} \times 840\,\text{s} = 1,260,000\,\text{J}

Electrical energy is the product of power in watts and time in seconds.

4

Convert energy from Joules to Megajoules

E = \frac{1,260,000}{10^6} = 1.26\,\text{MJ}

One Megajoule (

1\,\text{MJ}

) is equivalent to

10^6\,\text{Joules}

.

Key Concept

Electrical Energy Consumption

Question 198Question →

An electric water heater with an internal heating element of resistance $40\,\Omega$ is connected to a $200\,\text{V}$ mains power supply. If the heater is operated for $15\,\text{minutes}$ each day, what is the total electrical energy consumed by the heater over a period of $30\,\text{days}$ ?

7.5\,\text{kWh}

30.0\,\text{kWh}

450.0\,\text{kWh}

37.5\,\text{kWh}

Show answer & explanation

Answer: $7.5\,\text{kWh}$

Answer

The total electrical energy consumed over 30 days is

7.5\,\text{kWh}

.

The electrical power rating of the water heater is

P = \frac{V^2}{R} = \frac{200^2}{40} = 1000\,\text{W} = 1.0\,\text{kW}

. Operating for

15\,\text{minutes}

(

0.25\,\text{hours}

) daily for

30\,\text{days}

yields a total time of

7.5\,\text{hours}

. The total energy consumed is

1.0\,\text{kW} \times 7.5\,\text{h} = 7.5\,\text{kWh}

.

Step-by-Step Solution

1

Calculate the electric power rating of the heater

P = \frac{V^2}{R} = \frac{200^2}{40} = \frac{40000}{40} = 1000\,\text{W} = 1.0\,\text{kW}

Electric power dissipated in a resistance

R

connected across potential difference

V

is given by

P = \frac{V^2}{R}

.

2

Calculate the total operating time in hours

t = 30 \times \frac{15}{60} = 7.5\,\text{hours}

Commercial electrical energy is measured in kilowatt-hours (

\text{kWh}

), so time must be converted from minutes to hours.

3

Calculate total electrical energy consumed

E = P \times t = 1.0\,\text{kW} \times 7.5\,\text{h} = 7.5\,\text{kWh}

Electrical energy consumed is the product of power in kilowatts and total time in hours.

Key Concept

Calculation of commercial electrical energy consumption in kilowatt-hours using electric power and operating time.

Electricity and Magnetism

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