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Question
ε1 and ε2 are two batteries having emf of 34V and 10V respectively and internal resistance of 1Ω and 2Ω respectively. They are connected as shown in the figure below. Using Kirchhoff’s Laws of electrical networks, calculate the currents I1 and I2.
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Solution
By Kirchoff’s law of electrical networks consider closed loop ABEFA
`:. -4I_1 - 5(I_2 + I_2 ) - 7I_1 - I_1 + 34= 0`
`:. -4I_1 - 5I_1 - 5I_2 - 7I_1 - I_1 = -34`
`:. -17I_1 - 5I_2 = -34`
`:. 17I_1 + 5I_2 = 34` .....(1)
Consider closed loop BCDEB,
`4I_2 - 10 + 2I_2 + 7I_2 + 5(I_1 + I_2) = 0`
`:. 4I_2 + 2I_2 + 7I_2 + 5I_1 + 5I_2 = 10`
`18I_2 + 5I_1 = 10` ....(2)
Mutiply equation (1) by 5 and equation (2) by 17 and Substract
∴ `I_1 = 2A`
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