Question

The current across the 50 MV battery flowing through the circuit, where P Ω
(where P is the last 3 digits of your student ID), .[ id is 279 }100 Ω and 200 Ω resistors are
used. Solve the circuit and calculate the current across the battery. [CO2]

Answer 1

Answer:

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Answer 2

Given:

50 MV Battery

Three registers namely R1, R2 and R3 of Resistance 279 Ω, 100 Ω , 200 Ω

To Find:

Current across the battery.

Explanation:

The resistors can be connected in one of two ways

1. Series
2. Parallel$\frac{V}{I}$

Series :

In Series, the Resistance is Arithematically added to get the resultant resistance.

279 Ω + 100 Ω + 200 Ω = 579 Ω

According to Ohm's Law

$\frac{V}{I}$ = R

$\frac{10x^{6} }{I}$ = 579 Ω

I = 1727.11 A (Amps)

Parallel:

In Parallel, the reciprocal of the Resistance is added.

$\frac{1}{R} = \frac{1}{279} + \frac{1}{200} + \frac{1}{100}$

Taking LCM of all the denominators and making the fractions like

$\frac{1}{R} = \frac{200}{55800} + \frac{279}{55800} + \frac{558}{55800}$

R = $\frac{55800}{1037}$

R = 53.8 Ω

According to Ohm's Law

$\frac{V}{I}$ = R

$\frac{10x^{6} }{I}$ = 53.8 Ω

I = 18,587.36 A (Amps)

The Battery Across the Battery can be either 18,587.36 A ( Series )

or 1727.11 A ( Parallel ).