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In the circuit shown , find the current through resistors. |
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Answer» SOLUTION :Here `epsi_1 = 4V, epsi_2 = 8V, r_1 = 0.5 OMEGA ` and `r_2 = 1.0 Omega` Since TWO batteries are connected in opposite order, hence net EMF `epsi_(eq) = epsi_2 - epsi_1 = (8- 4) V= 4V` Now resistances `R_2=3.0Omega` and `R_3= 6.0 Omega` are joined in parallel, hence their effective resistance `R_23= (R_2 R_3)/(R_2 + R_3) = (3.0 xx 6.0)/(3.0 + 6.0) = 2.0 Omega` ` therefore ` Total resistance of entire circuit `R_(eq) =r_1 + r_2 +R_1 + R_23 = 0.5 + 1.0+ 4.5 + 2.0 = 8.0 Omega` ` therefore`Main circuit current `I= (epsi_(eq))/(R_(eq)) = (4V)/(8.0 Omega) = 0.5 A` ` therefore `Potential difference between points B and A, `V = IR_23 = 0.5 xx 0.2=1.0 V` ` therefore `Current flowing through `R_2 = 3.0 Omega , I_2= (V)/(R_2) = (1.0)/(3.0) = 0.33 A` and current flowing through `R_3 = 6.0 Omega , I_3 = (V)/(R_3) = (1.0)/(6.0) = 0.17 A ` |
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