Question

Two resistors, having resistances of 30 and 60 ohms are connected in parallel, and in series with them is a resistor having a resistance of 5 ohms. What p.d. must be applied to the whole circuit to cause a current of 4.8 Amps through the 5 ohm resistor?

Answer 1

Question#18474

Two resistors, having resistances of 30 and 60 ohms are connected in parallel, and in series with them is a resistor having a resistance of 5 ohms. What p.d. must be applied to the whole circuit to cause a current of 4.8 Amps through the 5 ohm resistor?

Solution:

Let:

$$R _ {1} = 3 0 \mathrm {o h m}$$ $$R _ {2} = 6 0 \mathrm {o h m}$$ $$R _ {3} = 5 \mathrm {o h m}$$ $$I = 4. 8 \mathrm {A m p}$$

$U - ?$

The current through the resistor R3 is equal to total current through circuit and according to the Ohm's Law is:

$I = \frac{U}{R_z}$ , were $R_{z}$ the total resistance of the circuit

$$R _ {z} = R _ {3} + \frac {R _ {1} * R _ {2}}{R _ {1} + R _ {2}} = 5 + \frac {3 0 * 6 0}{3 0 + 6 0} = 2 5 \mathrm {o h m}$$ $$U = I R _ {z} = 4. 8 * 2 5 = 1 2 0 \mathrm {V}$$

Answer: 120 V