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Find the magnitude and direction of the current flowing through the resistance `R` in the circuit shown in Fig. if the emf's of the sources are equal to `E_(1) = 1.5 V` and `E_(2) = 3.7 V` and the resistances are equal to `R_(1) = 10 Omega, R_(2) = 20 Omega , R = 5.0 Omega`. The internal resistances of the sources are neglible.

Text Solution

Verified by Experts

Make the current distribution as shown in the diagram,
Now, in the loop 12341, applying `-Delta varphi = 0`
`iR + i_(1) R_(1) + xi_(1) = 0` ...(1) ltbrltgt and in the loop 23562,
Solving (1) and (2), we obtains current through the resistance `R`,
`i = ((xi_(2) R_(1) - xi_(1) R_(2)))/(R R_(1) + R R_(2) + R_(1) R_(2)) = 0.02 A`
and it is directed from left to the right.
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Knowledge Check

  • In the circuit shown , the batteries have emf E_1 = E_2= 1V , E_3 = 2.5 V, and the resistance R_1 = 10Omega, R_2 = 20 Omega , Capacitance C = 10 muF . The charge on the left plate of the charge on the left plate of the capacitor C at steady state is

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