In the electric circuit shown each cell has an emf of, 2V and internal resistance of 1 Ω . The external resistance is 2 Ω . The value of the current I is: (in amperes):

Question

In the electric circuit shown each cell has an emf of, 2V and internal resistance of 1 Ω . The external resistance is 2 Ω . The value of the current I is: (in amperes): (A) 2 (B) 1.25 (C) 0.4 (D) 1.2

Tardigrade Admin · Accepted Answer

Currents=(net emf/net text resistance) Or I=(2+2+2/1+1+1+2)=(6/5)=1.2 A

Q. In the electric circuit shown each cell has an emf of, $2 V$ and internal resistance of $1 Ω$ . The external resistance is $2 Ω$ . The value of the current $I$ is: (in amperes):

2

1.25

0.4

1.2

0.8

$C u rre n t s = \frac{n e t e m f}{n e t res i s t an ce}$ Or
$I = \frac{2 + 2 + 2}{1 + 1 + 1 + 2} = \frac{6}{5} = 1.2 A$

Q. In the electric circuit shown each cell has an emf of, 2V and internal resistance of 1Ω . The external resistance is 2Ω . The value of the current I is: (in amperes):

2

1.25

0.4

1.2

0.8

Currents=netresistancenetemf​ Or I=1+1+1+22+2+2​=56​=1.2A

Q. In the electric circuit shown each cell has an emf of, $2 V$ and internal resistance of $1 Ω$ . The external resistance is $2 Ω$ . The value of the current $I$ is: (in amperes):

$C u rre n t s = \frac{n e t e m f}{n e t res i s t an ce}$ Or
$I = \frac{2 + 2 + 2}{1 + 1 + 1 + 2} = \frac{6}{5} = 1.2 A$