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# Two batteries with emf 12v and 13v are connected in parallel across a load resistor of 10ohm. Internal resistances are 1 and 2 ohm respectively. Range of the voltage across the load.

Arun
25763 Points
one year ago

Let VL represent the unknown load voltage. From KCL:

(13-VL)/2 + (12-VL)/1 = VL/10

Now multiply by 10 to remove fractions:

5(13-VL) + 10(12-VL) = VL

Expand terms:

65 -5VL + 120 - 10VL = VL

Then group like terms:

(65 + 120) + (-5VL - 10VL) = VL

And simplify:

185 - 15VL = VL

Bring all VL terms to right side:

185 = 16VL

and solve for VL:

VL = 185/16

Simulation gets the same value (rounded to two decimal places:

Khimraj
3007 Points
one year ago
from junction law we can write
(x-12)/1 + (x-13)/2 + (x-0)/0 = 0
x = 185/16
….....................................................
Ayush_Deep
120 Points
one year ago
Kindly draw the circuit according to the question . Then let the potential of lower polarity of battery be 0 and internal resistance connected to upper polarity.
Now apply KCL at the junction ,
(V--12)/1 + (V -- 13)/2 + (V--0)/10 =0
V = (120+65)/16
= 185/16
= 11.56
So correct answer will be option 2
Hope you understood