Question

Four resistances are joined together to form a
network as shown in the figure. The potential of the
Junction J will be
(1) 19.2 V
(3) 30.2 V
(2) 17.2 V
(4) 15.2 V​

Answer 1

Answer:

15.4 is the correct answer answer

Explanation:

Use kirchoffs voltage law you will get the answer

Mark it brainliest!!

Answer 2

Answer:

The voltage at the junction J is 19.2 V.

Explanation:

Assumption,

Let I₁, I₂, I₃, and I₄ be the current passing through the resistances 1 ohm, 2ohm, 3ohm, and 4ohm respectively.

Let the voltage at the junction J be 'V'.

Given,

The voltage at the terminals of 1ohm, 2ohm, 3ohm, and 4ohm resistances are 10V, 20V, 30V, and 40V respectively.

To find,

The voltage at the junction J (V)

Calculation,

Let the convention of current be:

The current I₁ is passing away from the junction J.

The current I₂ is passing towards the junction J.

The current I₃ is passing away from the junction J.

The current I₄ is passing towards the junction J.

From ohm's law

The current through the 1ohm resistor is I₁ = (V - 10)/1ohm

The current through the 2ohm resistor is I₂ = (20 - V)/2ohm

The current through the 3ohm resistor is I₃ = (V - 30)/3ohm

The current through the 4ohm resistor is I₄ = (40 - V)/4ohm

But As I₂ + I₄ = I₁ + I₃

$\implies \frac{20 - V}{2} + \frac{40 - V}{4} = \frac{V -10}{1} + \frac{V-30}{3} \\\implies \frac{40 - 2V}{4} +\frac{40 - V}{4} = \frac{3V- 30}{3} +\frac{V -30}{3}$

$\implies \frac{80 - 3V}{4} = \frac{4V - 60}{3} \\\impli$

⇒ 240 - 9V = 16V - 240

⇒ 480 = 25V

⇒ V = 19.2 V

Therefore, the voltage at the junction J is 19.2 V.

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