Let three batteries of emf E1=3 V,E2=2 V and E3=1 V and of internal resistances 1 Ω,2Ω and 1 Ω respectively constitute a circuit . They are connected in parallel as shown in the figure. The potential difference between points P and Q is "X" volts. The value of X is volts.

Question

Let three batteries of emf E1=3 V,E2=2 V and E3=1 V and of internal resistances 1 Ω,2Ω and 1 Ω respectively constitute a circuit . They are connected in parallel as shown in the figure. The potential difference between points P and Q is "X" volts. The value of X is volts. <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-8aconucseczcoyqv.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Ee f f=((3/1) + (2/2) + (1/1)/((1)1 + (1)2 + (1/1)) =2V

Q. Let three batteries of emf E1​=3V,E2​=2V and E3​=1V and of internal resistances 1Ω,2Ω and 1Ω respectively constitute a circuit . They are connected in parallel as shown in the figure. The potential difference between points P and Q is "X" volts. The value of X is __ volts.

Eeff​=(11​+21​+11​)13​+22​+11​​ =2V

$E_{e ff} = \frac{\frac{3}{1} + \frac{2}{2} + \frac{1}{1}}{( \frac{1}{1} + \frac{1}{2} + \frac{1}{1} )}$
$= 2 V$