Question

Two 12v batteries with internal resistance 0.2ohm and 0.25 ohm respectively are joined in parallel and a resistance of 1 ohm is placed across the terminals find the current supplied by each battery​

Answer 1

Answer:

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My question:-

Two batteries with emf 12 volt and 13 volt are connected in parallel across a load resister of the two batteries are 1 omega and 2 omega respectively.the voltage across the load lies between?

Answer 2

Therefore current supplied by each of the batteries is 6 Amp and 4.8 Amp.

Given:

The voltage of the batteries = V = 12 V

The internal resistances, R₁ = 0.2 ohm and R₂ = 0.25 ohm are  connected in parallel.

Resistance connected across the terminals R₃ = 1 ohm

To Find:

The current is supplied by each battery.

Solution:

The given question can be solved as shown below.

The circuit is formed with the given data as shown in the figure attached.

In the circuit,

Applying Kirchoff's Current Law across the circuit,

⇒ ( V - 12 ) / 0.2 + ( V - 12 ) / 0.25 + ( V - 0 ) / 1 = 0

⇒ { 5( V - 12 ) + 4( V - 12 ) + V } / 1 = 0

⇒ 5V - 60 + 4V - 48 + V = 0

⇒ 10V = 60 + 48 = 108

⇒ V = 10.8

Current passing through the first battery = I₁ = ( V - 12 ) / 0.2 = ( 10.8 - 12 ) / 0.2 = 1.2 / 0.2 = 6 Amp

Current passing through the second battery = I₂ = ( V - 12 ) / 0.25 = ( 10.8 - 12 ) / 0.25 = 1.2 / 0.25 = 4.8 Amp

Total current flowing across the circuit = I = I₁ + I₂ = 6 + 4.8 = 10.8 Amp

Therefore current supplied by each of the batteries is 6 Amp and 4.8 Amp.

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