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Asked by chamuthyag71 | 30 Jan, 2024, 05:08: PM

Part (a)

A1 = A3 = 1 A

i.e., Ammeter reading A1 and A3 are equal .

Ammeter reading  A1 shows the current enters the resitors network and

Ammeter reading  A3 shows the current exits the resitors network .

Both currents are one and the same.

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Part (b)

A2 = (1/4) A3

A3 is total current drawn from battery that equals 1 A .

Total current 1 A passes through 4 equal resistors that are connected parallel .

Hence current divides equally through each resistor .

Hence ammeter reading A2 = (1/4) A3

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Part (c)

V1 = 1.5 V

V1 is voltage across two 3Ω resistors that are connected parallel .

Equivalent resitance the resistors is 1.5Ω .

Since 1 A is passing through these resistors , we get 1.5 V across these resistors.

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Part (d)

Total resitance = 1.5Ω +  1 Ω + 0.75 Ω = 3.25 Ω

two parallel resistors each of resistance 3Ω gives equivalenet resistance 1.5 Ω .

three parallel resistors each of resistance 3Ω gives equivalenet resistance 1.0 Ω .

four parallel resistors each of resistance 3Ω gives equivalenet resistance 0.75 Ω .

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