ICSE Class 10 Physics 2016 Paper

A battery of emf $12\ \mathrm{V}$ and internal resistance $2\ \Omega$ is connected with two resistors $A$ and $B$ of resistance $4\ \Omega$ and $6\ \Omega$ respectively joined in series.

Find:

Question : 61 of 73

Marks: +1, -0

Current in the circuit.

Solution:

Given :

V = 12\ \mathrm{V}

, Resistance of

A = 4\ \Omega

, Resistance of

B = 6\ \Omega

, Internal battery resistance

= 2\ \Omega

\text{Total Resistance} = 4+6+2

= 12\ \Omega

\text{I (current in circuit)} = \frac{\text{e.m.f.}}{\text{Total resistance}}

= \frac{12}{12} = 1\ \text{amp.}

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