Question Video: Calculating the Resistance of the Galvanometer Coil Physics • Third Year of Secondary School

Video Transcript

The figure below shows a galvanometer scale when it is used to measure current and the scale when it is connected to a 4200-ohm multiplier resistor to measure the potential difference across a resistor. Calculate the resistance of the galvanometer coil. (A) 800 ohms, (B) 1250 ohms, (C) 2950 ohms, (D) 5000 ohms.

The question asks us to calculate the resistance of a galvanometer coil. The galvanometer is converted to a voltmeter by connection with a multiplier resistor of 4200-ohm resistance. Recall that a multiplier resistor must be connected in series with a galvanometer to convert the galvanometer to a voltmeter. Now, let’s also recall that a voltmeter must be connected in parallel to a circuit component, for example, a resistor 𝑅, to measure the potential difference across the component.

The question contains a diagram showing the range of values that the galvanometer can measure. The potential difference scale of the galvanometer has four marks other than the zero mark. The marks are equally spaced, and each mark measures a potential difference change of five volts. At full-scale deflection, that is, at this far end of the scale, the potential difference measured is therefore four times five volts, which is 20 volts. The current scale of the galvanometer uses the same marks as the potential difference scale. At full-scale deflection, the current measured is four milliamperes, or 0.004 amperes.

Next, let’s recall that there’s an equation for the multiplier resistance required to convert a galvanometer to a voltmeter. This equation says that the multiplier resistance 𝑅 𝑀 is equal to 𝑉 max over 𝐼 𝐺 minus 𝑅 𝐺, where 𝑉 max is the maximum potential difference that can be measured. 𝐼 𝐺 is the current through the galvanometer at this maximum potential difference. And 𝑅 𝐺 is the resistance of the galvanometer coil.

We’re trying to calculate the resistance of the galvanometer coil, so let’s rearrange the equation to make 𝑅 𝐺 the subject. We can add 𝑅 𝐺 to both sides of the equation. The positive and negative 𝑅 𝐺 terms then cancel on the right-hand side. We can then subtract 𝑅 𝑀 from both sides so that the positive and negative 𝑅 𝑀 terms on the left cancel out. We have then that 𝑅 𝐺 is equal to 𝑉 max over 𝐼 𝐺 minus 𝑅 𝑀. In this case, we have that the maximum potential difference 𝑉 max is equal to 20 volts.

We also know that the current 𝐼 𝐺 is 0.004 amperes. And we’re told that the multiplier resistance 𝑅 𝑀 is equal to 4200 ohms. Substituting these values into this equation, we have that 𝑅 𝐺 is equal to 20 volts over 0.004 amperes minus 4200 ohms. 20 volts over 0.004 amperes is equal to 5000 ohms. So, we have 5000 ohms minus 4200 ohms, which works out as 800 ohms. This matches the value given in option (A).

The correct answer is therefore given by option (A). The resistance of the galvanometer coil is 800 ohms.