Question

Two resistances R1 and R2 of 2 ohm and 3 ohm respectively are connected in parallel with a

1.5 volt battery and the battery current is found to be 0.75 A.

(i) Find the internal resistance of the battery.

(ii) Find the power developed across R1.​

Answer 1

Answer:

26.

(i) 0.8 ohm

(ii)current through R1 = (3/5) of 0.75 A

or, 0.45 A, power = I^2R1 = (0.45)^2

x 2 watt = 0.4 watt

Explanation:

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Answer 2

Given:

Two resistances 2 ohm and 3 ohm respectively are connected in parallel with a 1.5 volt battery and the battery current is found to be 0.75 A.

To find:

• Internal resistance of battery?
• Power developed across R1?

Calculation:

R1 and R2 are in parallel, so combination resistance will be :

$\therefore \: \dfrac{1}{R} = \dfrac{1}{2} + \dfrac{1}{3}$

$\implies \: \dfrac{1}{R} = \dfrac{3 + 2}{6}$

$\implies \: \dfrac{1}{R} = \dfrac{5}{6}$

$\implies \: R = 1.2 \: ohm$

Now, let internal resistance be r :

• This internal resistance is in series with the parallel combination of R1 and R2:

$\therefore \: i = \dfrac{V}{r + R}$

$\implies\: 0.75= \dfrac{1.5}{r + 1.2}$

$\implies\: r + 1.2 = 2$

$\implies\: r = 0.8 \: ohm$

So, internal resistance is 0.8 ohm.

• Potential drop across the parallel combination is : 0.75 × 1.2 = 0.9 Volt.

$\therefore \: P_{R1} = \dfrac{ {(0.9)}^{2} }{2}$

$\implies \: P_{R1} = \dfrac{0.81}{2}$

$\implies \: P_{R1} = 0.405 \: watt$

So, power developed across R1 is 0.405 watt.