Question

Find the power dissipation in resistance \[12\Omega \] in the circuit shown above figure.

A. \[27W\]
B. \[20W\]
C. \[50W\]
D. \[100W\]

Answer 1

Hint: In this problem we have to find the power dissipated in resistance \[12\Omega \]. We will first find the equivalent resistance between \[CD\] as they are in series connection. We will then find the equivalent resistance between \[AB\]. We will then find the power supplied into \[12\Omega \] resistance. We will use the fact that voltage across resistance connected in parallel is the same. And voltage across resistance connected in series divides.

Complete step by step answer:
This problem is based on resistance and ohm's law. Resistance is the property of any material through which it opposes the flow of current. It is denoted by symbol $R$. Ohm's law states that the current flowing through a conductor is directly proportional to the voltage across the conductor. Let as draw the given Circuit diagram with the resistance AB and CD is mentioned as follows

The resistance across \[CD\] are in series.
So the equivalent resistance across \[CD\] is given as \[{R_{CD}} = (7 + 5) = 35\Omega \].
Now the resistance across \[AB\] are in parallel.
Therefore equivalent resistance across \[AB\] is given by \[\dfrac{1}{{{R_{AB}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}\]
\[\dfrac{1}{{{R_{AB}}}} = \dfrac{1}{{12}} + \dfrac{1}{6} + \dfrac{1}{{12}} = \dfrac{1}{3}\]
Hence, \[{R_{AB}} = 3\]
Given voltage in the circuit \[ = \]\[60V\]
Voltage between \[AB\] \[ = \]\[{V_{AB}} = \dfrac{3}{{3 + 7}} \times 60\] on solving we have
Hence \[{V_{AB}} = 18V\]
Hence the voltage across resistor \[12\Omega \] is \[{V_{AB}} = 18V\]
Power dissipated in \[12\Omega \] resistance is \[P = \dfrac{{{V^2}}}{R}\]
Therefore, \[\ P = \dfrac{{{{18}^2}}}{{12}} = 27W\]
Hence, Power dissipated in \[12\Omega \] resistance is \[27W\].

Therefore, option \[A\] is correct.

Note: Resistance is the property of material, due to which it opposes the flow of current in the circuit. Equivalent resistance in series is given by \[{R_{eq}} = {R_1} + {R_2}\]. Equivalent resistance in parallel is given by \[\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}\]. Voltage across resistance in parallel is the same.Voltage across the resistor in series is different.