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.

Explanation

(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`