A 6 V cell with 0.5 Ω internal resistance, a 10 V cell with 1 Ω internal resistance and a 12 Ω external resistance are connected in parallel. The current (in ampere) through the 10 V cell is

Question

A 6 V cell with 0.5 Ω internal resistance, a 10 V cell with 1 Ω internal resistance and a 12 Ω external resistance are connected in parallel. The current (in ampere) through the 10 V cell is (A) 0.60 (B) 2.27 (C) 2.87 (D) 5.14

Tardigrade Admin · Accepted Answer

In closed loop ABGFEHA, 10-i2 × 1+i1 × 0.5-6=0 0.5 i1-i2=-4 (i) In closed loop B C D E B, (i1+i2) × 12+i2 × 1-10=0 12 i1+13 i2=10 ...(ii) From Eqs. (i) and (ii), we get i2=2.87 A

Q. A $6 V$ cell with $0.5 Ω$ internal resistance, a $10 V$ cell with $1 Ω$ internal resistance and a $12 Ω$ external resistance are connected in parallel. The current (in ampere) through the $10 V$ cell is

0.60

2.27

2.87

5.14

In closed loop $A BGFE H A$ ,
$10 - i_{2} \times 1 + i_{1} \times 0.5 - 6 = 0$
$0.5 i_{1} - i_{2} = - 4 (i)$
In closed loop $BC D EB$ ,
$(i_{1} + i_{2}) \times 12 + i_{2} \times 1 - 10 = 0$
$12 i_{1} + 13 i_{2} = 10 ... (ii)$
From Eqs. (i) and (ii), we get
$i_{2} = 2.87 A$

Q. A 6V cell with 0.5Ω internal resistance, a 10V cell with 1Ω internal resistance and a 12Ω external resistance are connected in parallel. The current (in ampere) through the 10V cell is