Solution Electrical Numerical 12

Numerical 12: Three batteries A, B and C have their negative terminals connected together, between the positive terminals of A and B there is a resistor of 0.5 ohm and between B and C three is a resistor of 0.3 ohm, i. Battery A 105 V, Internal resistance 0.25 ohm’ ii. Battery B 100 V, Internal resistance 0.2 ohm iii. Battery C 95 V, Internal resistance 0.25 ohm C. Determine the current values in the two resistors and the power dissipated by them.

Solution: 

From figure:

$\displaystyle \mathrm{105-100 = I_1\times 0.25 + I_1\times 0.5 + (I_1-I_2)\times 0.2}$ 

$\displaystyle \mathrm{100-95 = (I_2-I_1)\times 0.2 + I_2\times 0.3 + I_2\times 0.25}$ 

By solving the above two equations

$\displaystyle \mathrm{I_1\ =\ 7.063\ A}$ 

$\displaystyle \mathrm{I_2\ =\ 8.55\ A}$ 

Power dissipated in 0.5 ᘯ resistor = $\displaystyle \mathrm{I_1^2R_1\ =\ 7.06^2\times 0.5\ =\ 24.94\ W}$ 

Power dissipated in 0.3 ᘯ resistor = $\displaystyle \mathrm{I_2^2R_2\ =\ 8.535^2\times 0.3\ =\ 21.93\ W}$