A) 0.55W
B) 0.33 W
C) 0.25 W
D) 0.86 W
Correct Answer: C
Solution :
| [c] Resistors \[4\Omega ,6\Omega \]and \[12\Omega \] are connected in parallel, its equivalent resistance (R) is given by\[\frac{1}{R}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}\Rightarrow R=\frac{12}{6}=2\Omega \] |
| Again R is connected to 1.5 V battery whose internal resistance \[r=1\Omega .\] |
| Equivalent resistance now,\[R'=2\Omega +1\Omega =3\Omega \] |
| Current, \[{{I}_{total}}=\frac{V}{R}=\frac{1.5}{3}=\frac{1}{2}A\] |
| \[{{I}_{total}}=\frac{1}{2}=3x+2x+x=6x\]\[\Rightarrow \]\[x=\frac{1}{12}\] |
| \[\therefore \]Current through \[4\Omega \] resistor = 3x |
| \[=3\times \frac{1}{12}=\frac{1}{4}A\] |
| Therefore, rate of Joule heating in the \[4\Omega \]resistor\[={{I}^{2}}R={{\left( \frac{1}{4} \right)}^{2}}\times 4=\frac{1}{4}=0.25W\] |
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