### 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, letβs also recall that a
voltmeter must be connected in parallel to a circuit component, for example, a
resistor π
, 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, letβs recall that thereβs an
equation for the multiplier resistance required to convert a galvanometer to a
voltmeter. This equation says that the
multiplier resistance π
π is equal to π max over πΌ πΊ minus π
πΊ, where π max
is the maximum potential difference that can be measured. πΌ πΊ is the current through the
galvanometer at this maximum potential difference. And π
πΊ is the resistance of the
galvanometer coil.

Weβre trying to calculate the
resistance of the galvanometer coil, so letβs rearrange the equation to make π
πΊ
the subject. We can add π
πΊ to both sides of
the equation. The positive and negative π
πΊ
terms then cancel on the right-hand side. We can then subtract π
π from
both sides so that the positive and negative π
π terms on the left cancel out. We have then that π
πΊ is equal to
π max over πΌ πΊ minus π
π. In this case, we have that the
maximum potential difference π max is equal to 20 volts.

We also know that the current πΌ πΊ
is 0.004 amperes. And weβre told that the multiplier
resistance π
π is equal to 4200 ohms. Substituting these values into this
equation, we have that π
πΊ 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.