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Question

Physics Question on Current electricity

In the following network potential at O'O'

A

4 V

B

3 V

C

6 V

D

4.8 V

Answer

4.8 V

Explanation

Solution

Let the potential at OO is V0V_{0}.
Application of Kirchhoff's first law at junction OO gives
8V02=V044+V022\frac{8-V_{0}}{2} =\frac{V_{0}-4}{4}+\frac{V_{0}-2}{2}
=V04+2V044=\frac{V_{0}-4+2 V_{0}-4}{4}
4(8V0)2=V04+2V04\frac{4\left(8-V_{0}\right)}{2} =V_{0}-4+2 V_{0}-4
162V0=3V0816-2 V_{0} =3 V_{0}-8
16+8=5V016+8 =5 V_{0}
V0=245=4.8VV_{0} =\frac{24}{5}=4.8\, V