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For the circuit shown, with R(1)=1.0Omeg...

For the circuit shown, with `R_(1)=1.0Omega,R_(2)=2.0Omega,E_(1)=2V` and `E_(2)=E_(3)=4V`, the potential difference between the points `'a'` and `'b'` is approximately (in `V`):

Text Solution

Verified by Experts

The correct Answer is:
3.33
Let us assume `V_(b) = 0V and V_(a) = xV`
`i_(1) + i_(2) + i_(3) = 0`
`implies (x - E_1)/(2R_1) + (x = E_2)/(R_2) + (x - E_3)/(2R_1) = 0`
`=(x - 2)/2 + (x - 4)/(2) + (x - 4)/2 = 0`
`implies 3x - 10 = 0 implies x = 3.3 V`
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