ICSE Class 10 Physics 2016 Solved Paper

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

Find:

Question : 61 of 73

Marks: +1, -0

Current in the circuit.

Solution:

Given :

{\V}=12 {\V}

, Resistance of

{\A}=4 Ω

, Resistance of

{\B}=6 Ω

, Internal battery resistance

=2 Ω

\;\text " Total Resistance "\;\;=4+6+2

\;=12 Ω .

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

\;=\;{12}/{12}=1 {\amp.}

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