In a network with cells of negligible internal resistance, applying Kirchhoff's laws gives the equations $4I_1 - 3I_2 = 10$ and $4I_3 + 3I_2 = 5$ with $I_1 + I_2 = I_3$. The current $I_2$ equals:
A$-0.5$ A
B$+0.5$ A
C$2.12$ A
D$1.62$ A
Answer & Solution
Correct answer: A. $-0.5$ A
Substituting $I_3 = I_1 + I_2$ in the second equation: $4(I_1+I_2)+3I_2 = 5$ ⇒ $4I_1 + 7I_2 = 5$. Combining with $4I_1 - 3I_2 = 10$: subtracting gives $10I_2 = -5$ ⇒ $I_2 = -0.5$ A. The negative sign means $I_2$ flows opposite to the assumed direction.
Related questions
For a battery of emf ε and internal resistance r driving current I through external resistA galvanometer of resistance G is converted into a voltmeter reading full-scale voltage V A galvanometer of resistance G shows full-scale deflection at current I_g. To convert it iThe relation between current density j, drift velocity v_d, number density n and charge e A wire has a resistance R. It is stretched uniformly so that its length becomes 2L. The neKirchhoff current law at a junction is a statement ofTwo cells of emf ε and internal resistance r each are connected in parallel. The equivalenA potentiometer of length L is used to compare emfs. If the balance lengths for cells of e