Two resistance R1 and R2 are connected in series with a battery and an ammeter of negligible resistance. If a low resistance volt-meter is connected across R1 the ammeter reading.
Answer & explanation
Correct answer: option 1
Initially the resistance of circuit is R1 + R2. When a low resistance voltmeter of resistance Rv is connected in parallel across R1 the effective resistance becomes
$ R_2 + \frac{R_1 R_v}{R_1 + R_v} < R_1 + R_2$
When resistance of circuit decreases, current is the circuit increases.