A) \[5.76\]
B) \[1.44\]
C) \[2.88\]
D) \[2.02\]
Correct Answer: C
Solution :
: As per question, Heat produced in \[5\Omega \]. in time t is \[4.05=I_{1}^{2}5\] \[{{I}_{1}}=\sqrt{\frac{4.05}{5}}=0.9A\] Potential difference across B and C is \[=5{{I}_{1}}=5\times 0.9V=4.5V\] \[{{I}_{2}}=\frac{4.5V}{15\Omega }=0.3A\] \[\therefore \] \[I={{I}_{1}}+{{I}_{2}}=0.9A+0.3A=1.2A\] Heat produced in \[2\Omega \] in same time interval t is \[={{(1.2A)}^{2}}(2\Omega )=2.88J\]You need to login to perform this action.
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