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Difficulty: HardElectric Circuits and Measuring Instruments

A galvanometer with an internal resistance of 100 Ω100\ \Omega gives a full-scale deflection for a current of 1.0 mA1.0\text{ mA}. It is converted into an ammeter capable of measuring up to 1.0 A1.0\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 Ω4.40\ \Omega and a cell of electromotive force 3.0 V3.0\text{ V} having an internal resistance of 0.50 Ω0.50\ \Omega. What is the current reading indicated by the ammeter?

  1. 0.60 A0.60\text{ A}Answer
  2. B
    0.67 A0.67\text{ A}
  3. C
    0.68 A0.68\text{ A}
  4. D
    0.03 A0.03\text{ A}

Answer

The ammeter will indicate a current of 0.60 A0.60\text{ A}.
To find the ammeter reading, first calculate the effective ammeter resistance RA=IgGI=0.001×1001.0=0.10 ΩR_A = \frac{I_g G}{I} = \frac{0.001 \times 100}{1.0} = 0.10\ \Omega. The total circuit resistance is Rtotal=RA+R+r=0.10+4.40+0.50=5.00 ΩR_{\text{total}} = R_A + R + r = 0.10 + 4.40 + 0.50 = 5.00\ \Omega. Applying Ohm's law I=ERtotal=3.05.00=0.60 AI = \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 (RAR_A).
RA=Ig×GImax=0.001 A×100 Ω1.0 A=0.10 ΩR_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 (RtotalR_{\text{total}}) of the complete series circuit.
Rtotal=RA+R+r=0.10 Ω+4.40 Ω+0.50 Ω=5.00 ΩR_{\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 (II) using Ohm's law for a complete circuit.
I=ERtotal=3.0 V5.00 Ω=0.60 AI = \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
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Topics

Electric Circuits and Measuring Instruments

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