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In the circuit shown, i. VA - VB = ……....

In the circuit shown,
i. `V_A - V_B = ……..
(ii) I_1 = …….
(iii) I_2 ……..
iv. I = ……..

Text Solution

Verified by Experts

The given cirucit may be replaced by an equivalent battery of emf `epsilon` and resistanace r as shown in fig 5.102

`epsilon = ((4)/(4) + ((-6)/(12)))/((1)/(4)+(1)/(12))=(3)/(2)V`
and `r = (1)/((1)/(4)+(1)/(12))=3Omega`

Thus, current through the external resistor is (From A to B)
`I = (E )/(r+R) = (3//2)/(3+6) = (1)/(6)A`
For potential difference `V_A - V_B:`
`V_A - 6xx(1)/(6) = V_B`
or `V_A - V_B = 6xx(1)/(6) = 1V`
For upper branch:
`V_A - I_1 xx12+6 = V_B`
or `(V_A - V_B)+6 = I_1 x12 or 1+6 = I_1xx12 or I_1xx12 or I_1 = (7)/(12)A`
For lower branch:
`V_A +I_2xx4-4 = V_B`
or ` (V_B - V_B)- 4 = I_2xx4or I_2 = (3)/(4)A`
Method 2. Instead of making equivalent battery of two
bracnhes, we can make equivalent battery fo all three branches.
in the middle bracnh, we can assume a battery of zero emf
connected with resistance `6Omega.` The given ciruit may be
replaced by an equivalent battery of emf `epsilon` and resistance r as
shwon in fig 5.103
`epsilon = ((4)/(4)+(-(6)/(12))+(0)/(6))/((1)/(4)+(1)/(12)+(1)/(6))=1V`
and `r = (1)/((1)/(4)+(1)/(12)+(1)/(6))= 2Omega`
This battery is an open cirucit battery, hence potential
differecne across each branch will be 1V. Now we will get the
saem result as we have in previous method.
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