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Three resistors of 6.0 Ω, 2.0 Ω and 4.0 Ω are joined to an ammeter A and a cell of e.m.f. 6.0 V as shown in figure. Calculate —

(a) the effective resistance of the circuit, and

(b) the reading of ammeter.

Three resistors of 6.0 Ω, 2.0 Ω and 4.0 Ω are joined to an ammeter A and a cell of e.m.f. 6.0 V as shown in figure. Calculate the effective resistance of the circuit, and the reading of ammeter. Current Electricity, Concise Physics Solutions ICSE Class 10.

Current Electricity

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Answer

In the circuit, there are two parts. In the first part, resistors of 2.0 and 4.0 Ω are connected in series. If the equivalent resistance of this part is Rs then

Rs = 2 + 4 = 6 Ω

In the second part, Rs = 6.0 and resistor of 6.0 Ω are connected in parallel. If the equivalent resistance of this part is Rp then

1Rp=16+161Rp=1+161Rp=26Rp=62Rp=3.0Ω\dfrac{1}{Rp} = \dfrac{1}{6} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{Rp} = \dfrac{1 + 1}{6} \\[0.5em] \dfrac{1}{Rp} = \dfrac{2}{6} \\[0.5em] Rp = \dfrac{6}{2} \\[0.5em] R_p = 3.0 Ω \\[0.5em]

Hence, the effective resistance of the circuit = 3 Ω

(b) The reading of ammeter = ?

R = 3 Ω

V = 6.0 V

Using Ohm's law,

V = IR

Substituting the values in the formula above we get,

6 = I x 3
⇒ I = 6 / 3 = 2 A

Hence, the reading of ammeter = 2 A

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