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Q.
In the circuit shown, the current through 8 ohm is same before and after connecting $\varepsilon$. The value of $\varepsilon$ is
Current Electricity
Solution:
Before connecting $\varepsilon$, the circuit diagram is as shown in the figure.
The equivalent resistance of the given circuit is
$R_{ eq }=6\, \Omega+8\, \Omega+10\, \Omega=24\, \Omega$
Current in the circuit, $I=\frac{12 V }{24\, \Omega}=\frac{1}{2} A$
Before connecting $\varepsilon$, the current through $8 \Omega$ is
$I=\frac{1}{2} A$
After connecting $\varepsilon$, the current through $8\, \Omega$ is also
$I=\frac{1}{2} A;$
$\therefore \varepsilon=\frac{1}{2} A \times 8\, \Omega=4\, V$
