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In the circuit shown in fig. E(1)=3" vol...

In the circuit shown in fig. `E_(1)=3" volt",E_(2)=2" volt",E_(3)=1" vole and "R=r_(1)=r_(2)=r_(3)=1`ohm.
(i) Find potential difference in Volt between the points A and B with A & B unconnected.
(ii) If `r_(2)` is short circuited and the point A is coneted to point B through a zero resistance wire, find the current through R in ampere.

Text Solution

Verified by Experts

Applying Kirchoff.s loop law to mesh PLMQP and PLMQONP in the figure shown below, we  have
`i_(1)r_(1)+i_(2)r_(2)=E_(1)-E_(2)ori,+i_(2)=1ldots(i)`
`i_(1)r_(1)+i_(3)r_(3)=E_(1)-E_(3)ori-(1)+i-(3)=2ldots`(ii)`AtP,i_(2)+i_(3)=i_(1)`ldots(iii)
On solving (i), (ii) and (iii) `i_(1) = 1 amp. i_(2) = 0 amp. i_(3) = 1 amp`

Since no current is drawn along the branch APtherefore`V_(AB) = V_(PQ)`
Potential difference across PQ, ,br>`V_(PQ) = E_(1) -i_(1)r_(1)=2Votl`
(b) The figure shows the circuit when point A is connected to point B and `r_(2)` is short-circuited.Applying Kirchoff.s junction rule at P, we get`i = i_(1) + i_(2) +i_(3)`ldots(iv)
Applying Kirchoff.s law to mesh ABMLA `i-(1)r_(1) = E_(1)-E_(2) or i_(1) = 1 amp`
Applying Kirchoff.s law to mesh ANOQML`i_1r_(1)-i_(3)r_(3)=E_(1)-E_(3)ori_(1)+I_(2)+I_(3)=2ldots(v)`
From above equations `I_(1) = 1 amp, i_(2) = 2 amp, i_(3) = -1 amp`
.lamp (direction of current is opposite)
So, current through resistor R will be `l = I_(1)+l_(2)+ I_(3) = 2 amp`
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