Resistors in Parallel and Series problems.?

So the potential difference across the 10-ohm resistor is 7.40 V, and that's the potential difference across the battery, too.

Total resistance in Part 2 is 14.4 ohms; the potential difference across the battery is (1.6 A)(14.4 ohm) = around 23 V but use a calculator.

Part 1 of 2)

10 Ω resistor and a 5.00 Ω resistor are connected in parallel to a battery, and the current in the 10 Ω resistor is found to be 0.740 A.Find the potential difference V across the battery.

V = R*I = 0.74*10 = 7.4 V

Part 2 of 2)

R1 = 9.2 Ω resistor and a R2 = 5.2 Ω resistor are connected in series to a battery, and the current through the 9.2 Ω resistor is I' = 1.6 A.Find the potential difference V' across the battery.

V' = (R1+R2)*I' = (9.2+5.2)*1.6 = 23.0 V 

1.  E=10*.740=7.4V

2.  E=9.2+5.2=14.4Ω*1.6A=23.04V

E = IR = 0.74•10 = 7.4 volts

E = IR = 1.6(9.2+5.2) = 23 volts