A) \[30\,\Omega \]
B) \[8\,\Omega \]
C) \[10\,\Omega \]
D) \[40\,\Omega \]
Correct Answer: B
Solution :
The circuit can be shown as given below The equivalent resistance between D and C \[{{R}_{DC}}=\frac{15\times (15+15)}{15+(15+15)}=\frac{15\times 30}{15+30}\] \[=\frac{15\times 30}{45}=10\,\Omega \] Now, between A and B, the resistance of upper part ADCB, \[{{R}_{1}}=15+10+15=400\] Between A and B, the resistance of middle part AOB \[{{R}_{1}}=15+10+15=400\] Therefore, equivalent resistance between A and B \[\frac{1}{R'}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}=\frac{1}{{{R}_{3}}}=\frac{1}{40}+\frac{1}{30}+\frac{1}{15}\] \[=\frac{3+4+8}{120}=\frac{15}{120}\Rightarrow \,\,\,\,R'=\frac{120}{15}=8\,\,\Omega \]You need to login to perform this action.
You will be redirected in
3 sec