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In the circuit shown E, F, G and H are c...

In the circuit shown `E, F, G` and `H` are cells of emf `2V, 1V, 3V` and `1V` respectively and their internal resistance are `2Omega, 1Omega, 3Omega` and `1Omega` respectively.

Text Solution

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(i) `2/13 V, (ii) V_G=21/13V, V_H=9/13V`
Applying kirchoff's second law in loop BADB, `2-2j_1-j_1-1-2(j_1-j_2)=0`
Similarly applying kirchoff's second law in loop BDCB
`2(j_1-j_2)+3-3j_2-j_2-1=0`………(ii)
Solving Eqs. (i) and (ii) we get
`j_1=5/13,j_2=6/13 and j_1-j_2=-1/13`
(i) Potential difference between B and D,
`V_B+2(j_1-j_2)=V_D`
`therefore V_B-V_D=-2(j_1-j_2)=2/13 V`
(ii) `V_G=E_G-j_2r_G=3-6/13 times 3=21/13 V, V_H=E_H+j_2r_h=1+6/13 times 1=19/13V`
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