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NeTakaya
2 years ago
10

A battery of emf 20v and internal resistance 1 ohms connected in series with an external resistor4 ohms what is

Physics
1 answer:
Harlamova29_29 [7]2 years ago
8 0

Answer:

The total rate at which electrical energy is dissipated by the two resistors is 48 W.

The given parameters;

First resistor, R₁ = 25 ohm

Second resistor, R₂ = 15 ohm

power dissipated in the first resistor, P₁ = 36 W

In series circuit arrangement, the current flowing in each resistor is the same.

The current flowing in the first resistor is calculated as follows;

P = I²R₁

The equivalent resistance of the series arrangement is calculated as;

R = R₁ + R₂

R = 25 + 15

R = 40 ohms

The total rate at which electrical energy is dissipated by the two resistors;

P = I²R

P = 1.2(40)

P = 48 W

Thus, the total rate at which electrical energy is dissipated by the two resistors is 48 W.

Explanation:

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