Q4. Assertion: an ammeter is made by joining a small resistance in parallel with a galvanometer. Reason: most of the current then passes through the small resistance, and the meter’s total resistance is small. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: the shunt protects the coil and keeps the ammeter’s resistance close to rs, so it hardly changes the current it measures (NCERT page 130).
Q5. A galvanometer of resistance 20 Ω reads full scale at 5.0 mA. Find the shunt needed to make it read up to 2.0 A, and the resistance of the resulting ammeter.
rs = IgRG/(I − Ig) = 0.0050 × 20/1.995 ≈ 0.050 Ω. The ammeter’s resistance is 20 × 0.050/(20 + 0.050) ≈ 0.050 Ω (reasoned from Ohm’s law; NCERT page 130).
Q6. A voltmeter made from a galvanometer (RG = 100 Ω, full scale at 1.0 mA) reads up to 5.0 V. What resistance must be added in series to extend its range to 20 V?
For 5.0 V the total is 5.0/0.0010 = 5000 Ω; for 20 V it must be 20 000 Ω. So 15 000 Ω more is added in series (reasoned from Ohm’s law; NCERT page 131).
Q7. The number of turns of a galvanometer coil is doubled, with the same wire gauge and area per turn, while B and k stay fixed, so that twice the wire length doubles its resistance. What happens to (a) its current sensitivity, (b) its voltage sensitivity?
(a) It doubles, since φ/I = NAB/k. (b) It is unchanged, since φ/V = NAB/(kRG) and N and RG both double (NCERT page 131).