You visited us 1 times! Enjoying our articles? Unlock Full Access!

Heating Effect of Electric Current

In the circui...

Question

In the circuit above, $R_{1} = 2 R_{3}$ and $R_{3} = 2 R_{2}$ . If the source voltage is $V_{1}$ , find the power dissipated in $R_{1}$ .

\frac{(V_{1})^{2}}{7 R_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{(V_{1})^{2}}{14 R_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{(3 V_{1})^{2}}{7 R_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{(V_{1})^{2}}{49 R_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{(3 V_{1})^{2}}{49 R_{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is E $\frac{(3 V_{1})^{2}}{49 R_{2}}$
Given : $R_{1} = 2 R_{3}$ $R_{3} = 2 R_{2}$
$⟹$ $R_{1} = 2 (2 R_{2}) = 4 R_{2}$

Equivalent resistance of $R_{2}$ and $R_{3}$ $r = \frac{R_{2} \times R_{3}}{R_{2} + R_{3}} = \frac{R_{2} \times 2 R_{2}}{R_{2} + 2 R_{2}} = \frac{2}{3} R_{2}$

$∴$ Equivalent resistance of whole circuit $R_{e q} = r + R_{1} = \frac{2}{3} R_{2} + 4 R_{2} = \frac{14}{3} R_{2}$

Voltage across $R_{1}$ $V^{'} = \frac{R_{1}}{R_{e q}} V_{1} = \frac{4 R_{2}}{(14 / 3) R_{2}} V_{1} = \frac{6}{7} V_{1}$

$∴$ Power dissipated in $R_{1}$ $P^{'} = \frac{(V^{'})^{2}}{R_{1}} = \frac{(\frac{6}{7} V_{1})^{2}}{4 R_{2}} = \frac{(3 V_{1})^{2}}{49 R_{2}}$

Suggest Corrections

Similar questions

V_{1}

V_{2}

‘ m^{'}

V_{1}

V_{2}

x_{1}

x_{2}

v_{1}

v_{2}

The refracting angle of a prism is A, and refractive index of the material of the prism is cot $(\frac{A}{2})$ . The angle of minimum deviation is :

r_{1}, r_{2}

r_{3}

x^{3} - 2 x^{2} + 4 x + 5074 = 0

(r_{1} + 2) (r_{2} + 2) (r_{3} + 2)

Join BYJU'S Learning Program