Question

Figure shows a part of an electric circuit. The potentials at points a, b and c are 30 V, 12 V and 2 V respectively. The current through the 20 Ω resistor will be.

पर्याय

• 0.4 A

• 0.2 A

• 0.6 A

• 1.0 A

Answer 1

0.4 A

Explanation:

Given: Potential at point a (Va) = 30 V

Potential at point b (V_b) = 12 V

Potential at point c (V_c) = 2 V

Let the potential at the central junction point be Vj.

According to Kirchhoff’s Current Law, the sum of currents entering the junction must equal the sum of currents leaving it:

I_a = I_b + I_c    ...(i)

Using Ohm’s Law :

I = `(Delta V)/R`

Substituting this value in equation (i), we get,

`(V_a - V_j)/10 = (V_j - V_b)/20 + (V_j - V_c)/30`

⇒ `(30 - V_j)/10 = (V_j - V_b)/20 + (V_j - V_c)/30`    ...[Multiply the entire equation by the LCM of denominators (60) to clear them]

⇒ 6(30 − V_j) = 3(V_j − 12) + 2(V_j − 2)

⇒ 180 − 6V_j = 3V_j − 36 + 2V_j − 4

⇒ 180 − 6V_j = 5V_j − 40

⇒ 180 + 40 = 5V_j + 6V_j

⇒ 220 = 11V_j

⇒ V_j = `220/11`

⇒ V_j = 20 V

Now, use the junction potential to find the specific current flowing toward the point.

I_b = `(V_j - V_b)/20`

= `(20 - 12)/20`

= `8/20`

= 0.4 A