Exam

JIPMER

Year

2011

Subject

Physics

Chapter

Current Electricity

JIPMER 2011 · Physics · Current Electricity

In the circuit shown the value of I in ampere is

In the circuit shown the value of $I$ in ampere is

Step-by-step solution

So net resistance, $R=2.4+1.6=4.0 \Omega$ Therefore, current from the battery, $i=\frac{V}{R}=\frac{4}{4}=1 \mathrm{~A}$ Now from the circuit (b), $$\begin{aligned} & 4 I^{\prime}=6 I \\ \Rightarrow \quad & I^{\prime}=\frac{3}{2} I\end{aligned}$$ But, $\quad i=I+I^{\prime}=I+\frac{3}{2} I=\frac{5}{2} I$$\begin{array}{lc}\therefore & 1=\frac{5}{2} I \\ \Rightarrow & I=\frac{2}{5}=0.4 \mathrm{~A}\end{array}$$