Class 12th Physics 2017 Set1 Delhi Board Paper Solution
The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r as shown in the figure. Obtain the expression for
(i) the current draws from the cell and
(ii) the power consumed in the network.
Apply Horizontal symmetry to get current,
r = resistance of the circuit and internal resistance of r
1 / RI = 1/r + 1/2r = 2+1 / 2r = 3/2 r
Or RI = 2/3r
Circuit I and II are same RII = 2/3r
Combining resistance 1/R = 1/RI + 1/RII = 1/2/3r + 1/2/3r = 2 x 1/2/3r
⇒ 1/R = 3/r
⇒ R = r/3
This circuit is in series with internal resistance,
Resultant resistance : r + r/3 = 4r / 3 â¦.
Current drawn from the cell = I = 3E /4r
and power consumed by the network P = I2R = (3E /4r )2 x 4r / 3
.= 3E2/4 r
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