Current Electricity Ques 48
- In the circuit shown in figure $E, F, G, H$ are cells of emf $2, 1 , 3$ and $1 $ $V$ respectively, and their internal resistances are $2, 1, 3$ and $1$ $ \Omega$ respectively. Calculate
$(1984,4 M)$
(a) the potential difference between $B$ and $D$ and
(b) the potential difference across the terminals of each cells $G$ and $H$.
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Answer:
Correct Answer: 48.$(a) \frac{2}{13}$ $ V ,(b) \frac{21}{13}$ $ V ,\frac{19}{13}$ $ V $
Solution:
Formula:
- Applying Kirchhoff’s second law in loop $B A D B$
$48-2 i _1-i _1-1-2\left(i _1-i _2\right)=0 $ $\quad$ …….(i)
Similarly applying Kirchhoff’s second law in loop $B D C B$
$ 2\left(i _1-i _2\right)+3-3 i _2-i _2-1=0 $ $\quad$ …….(ii)
Solving Eqs. (i) and (ii), we get
(a) Potential difference between $B$ and $D$.
$ \begin{aligned} V _B+2\left(i _1-i _2\right) & =V _D \\ \therefore \quad V _B-V _D=-2\left(i _1-i _2\right) & =\frac{2}{13} V \end{aligned} $
(b)
$ \begin{gathered} V _G=E _G-i _2 r _G=3-\frac{6}{13} \times 3=\frac{21}{13} V \\ V _H=E _H+i _2 r _H=1+\frac{6}{13} \times 1=\frac{19}{13} V \end{gathered} $
BETA
