Find the equivalent resistance of circuit shown in the figure. Find also the current in $R_{3}$ and $R_{4}$ .

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Solution

equivalent resistance of circuit shown in fig is,
$R_{e q} = \frac{4 \times 4}{4 + 4} + 1 + \frac{12 \times 6}{12 + 6} + 1$

$∴ R_{e q} = 8 Ω$

Current in $R_{3}$ is given by,
$I = I_{R_{3}} = \frac{V}{R_{e q}} = \frac{16}{8} = 2 A$

Current in $R_{4}$ is given by,
$I_{R_{4}} = I \times \frac{R_{5}}{R_{4} + R_{5}} = 2 \times \frac{6}{12 + 6}$

$∴ I_{R_{4}} = 0.66 A$

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Similar questions

R_{3}

(R_{1} = 15 Ω, R_{2} = 15 Ω, R_{3} = 30 Ω, R_{4} = 35 Ω)

In the circuit given E = 6.0 V, $R_{1} = 100 Ω, R_{2} = R_{3} = 50 Ω, R_{4} = 75 Ω$ . The equivalent resistance of the circuit, in $Ω$ , is

\frac{R_{2}}{R_{4}}

R_{3}

R_{2}

R_{4} = 2 Ω

R_{3} = 4 Ω

E = 6.0 V, R_{1} = 100 Ω, R_{2} = R_{3} = 50 Ω

R_{4} = 75 Ω

(in Ω)

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