Q4. Assertion: in the half-deflection method the shunt resistance S that halves the deflection is nearly equal to the galvanometer’s resistance G. Reason: when R is much larger than G, closing the shunt hardly changes the total current, so the current divides almost equally between G and S. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: with the total current nearly unchanged, half of it goes through G only if S ≈ G; the exact result is G = RS/(R − S).
Q5. With a 1.5 V cell and R = 2950 Ω a galvanometer shows 30 divisions; a shunt of 50 Ω halves the deflection. Find G and the figure of merit.
G = 2950 × 50/2900 ≈ 50.9 Ω. Then k = ε/[(R + G)θ] = 1.5/(3000.9 × 30) ≈ 1.67 × 10−5 A per division, about 17 µA per division.
Q6. A student forgets the cell’s internal resistance r. How does it change the figure of merit worked out from k = ε/[(R + G)θ]?
The true current is ε/(R + G + r), so the formula slightly overestimates k. When R is a few kilohms and the chosen source has r of only a few ohms, the error is a fraction of a per cent; this is not guaranteed for every cell. G is affected far less: r only adds to R, so G = (R + r)S/(R + r − S), which for Q5 with r = 5 Ω differs from RS/(R − S) by under 0.01 % (reasoned from Ohm’s law).
Q7. The galvanometer of Q5 has 30 divisions on each side. What resistance must be added in series to make it a voltmeter reading up to 3.0 V?
The full-scale current is 30k ≈ 5.0 × 10−4 A, so the total resistance must be about 3.0/(5.0 × 10−4) ≈ 6000 Ω, and the series resistance about 6000 − 51 ≈ 5950 Ω, or 5.95 kΩ (reasoned from Ohm’s law).