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In the circuit shown in Fig. the emf of ...

In the circuit shown in Fig. the emf of the sources is equal to `xi = 5.0 V` and the resistances are equal to `R_(1) = 4.0 Omega` and `R_(2) = 6.0 Omega`. The internal resistance of the source equals `R = 1.10 Omega`. Find the currents flowing through the resistances `R_(1)` and `R_(2)`.

A

1A,1A

B

1.2A,0.8A

C

1.8A,1.3A

D

1.3A,1.9A

Text Solution

Verified by Experts

Using Kirchoff’s loop rule in loops (1) and (2), we have `E-R(i_1+i_2)-i_1R_1=0 and –i_1R_1+i_2R_2=0` . After solving equations, and substituting the known values, we get , `i_1=1.2A and i_2=0.8A`
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