Current Electricity 2 Question 17
17. In the circuit shown in the figure, the current through
$(1998,2 M)$
(a) the $3 \Omega$ resistor is $0.50 A$
(b) the $3 \Omega$ resistor is $0.25 A$
(c) the $4 \Omega$ resistor is $0.50 A$
(d) the $4 \Omega$ resistor is $0.25 A$
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Solution:
- Net resistance of the circuit is $9 \Omega$.
$\therefore$ Current drawn from the battery,
$$ i=\frac{9}{9}=1 A=\text { current through } 3 \Omega \text { resistor } $$
Potential difference between $A$ and $B$ is
$$ \begin{aligned} & & V _A-V _B & =9-1(3+2)=4 V=8 i _1 \\ \therefore & & i _1 & =0.5 A \\ \therefore & & i _2 & =1-i _1=0.5 A \end{aligned} $$
Similarly, potential difference between $C$ and $D$
$$ \begin{aligned} & V _C-V _D=\left(V _A-V _B\right)-i _2(2+2) \\ & =4-4 i _2=4-4(0.5)=2 V=8 i _3 \\ & \therefore \quad i _3=0.25 A \\ & \text { Therefore, } \quad i _4=i _2-i _3=0.5-0.25 \\ & i _4=0.25 A \end{aligned} $$
