1.

A battery 10 V and 0.5 Omega internal resistance is connected to a battery of 12 V and 0.8 Omega internal resistance and one terminal of battery is connected to a 20 Omega resistance, then the current flow in 20 Omega resistance is

Answer»

0.3023 A
0.8034 A
0.5303 A
1.238 A

Solution :According to KCL, the flow of current at node F and C is marked in the circuit.

Now, in the PART FABCF of the circuit
Applying KVL, `0.5i_(1)+0.8(i_(1)+i_(2))=12-10`
`0.5i_(1)+0.8i_(1)+0.8 i_(2)=2`
`RARR "" 1.3i_(1)+0.8i_(2)=2`.....(i)
SIMILARLY, applying KVL in FCDEF
`-0.8(i_(1)+i_(2))-20i_(2)=-12`
`-0.8i_(1)-0.8i_(2)-20i_(2)=-12`
`0.8i_(1)+20.8i_(2)=12`.....(ii)
On SOLVING Eqs. (i) and (ii), `i_(1)=1.212 A` and `i_(2)=0.5303 A`


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