An electric circuit consists of a voltage source with an electromotive force (e.m.f.) of 6… — JAMB UTME

An electric circuit consists of a voltage source with an electromotive force (e.m.f.) of $6.0\text{ V}$ and an internal resistance of $1.0\text{ }\Omega$ connected across two identical $6.0\text{ }\Omega$ resistors arranged in parallel. If an ammeter of negligible resistance is connected in series with the voltage source, what current will be displayed on the ammeter?

$1.5\text{ A}$ Answer
B
$2.0\text{ A}$
C
$0.75\text{ A}$
D
$0.46\text{ A}$

Answer

The ammeter will record a total main circuit current of

1.5\text{ A}

The correct response demonstrates proper circuit reduction. First, combining the two

6.0\text{ }\Omega

parallel resistors gives an equivalent resistance of

3.0\text{ }\Omega

. Second, adding the

1.0\text{ }\Omega

internal resistance yields a total resistance of

4.0\text{ }\Omega

. Applying Ohm's law (

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

) gives

I = \frac{6.0\text{ V}}{4.0\text{ }\Omega} = 1.5\text{ A}

, which is the total current measured by an ammeter placed in the main circuit.

Step-by-Step Solution

Calculate the equivalent external resistance (

R_p

) of the two parallel resistors.

R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = \frac{36.0}{12.0} = 3.0\text{ }\Omega

Resistors in parallel combine according to

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

Determine the total resistance (

R_{\text{total}}

) of the complete circuit, including internal resistance.

R_{\text{total}} = R_p + r = 3.0\text{ }\Omega + 1.0\text{ }\Omega = 4.0\text{ }\Omega

Internal resistance acts in series with the equivalent external load resistance.

Apply Ohm's law for a complete circuit to find the total current (

I

I = \frac{E}{R_{\text{total}}} = \frac{6.0\text{ V}}{4.0\text{ }\Omega} = 1.5\text{ A}

The total current supplied by the source equals the electromotive force divided by the total circuit resistance.

Key Concept

Terminal Circuit Current with Internal Resistance and Parallel Resistor Combinations

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic