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Study the electric circuit of Fig. 12.30 and find (i) the current flowing in the circuit and (ii) the potential difference across 10 Omega resistor. |
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Answer» Solution :(i) Here V=3V, `R_1=10 OMEGA and R_2=20 Omega` Since `R_1 and R_2` are connected in series the effective resistance of the circuit `R=R_1+R_2=10+20=30 Omega` `THEREFORE` Current FLOWING in the circuit `I=V/R=3/30=0.1A` (ii) The potential difference across `R_1=10 Omega` resistor is `V_1=IR_1=0.1 times 10=1.0 V` 
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