Four Equal Resistances Dissipated 5W

Four equal resistances dissipated 5W of power together when connected in series to a battery of negligible internal resistance. The total power dissipated in these resistances when connected in parallel across the same battery would be 80 W.

Solution

Assume voltage of battery terminals =E

Let us assume resistance of all four resistors as R

When resistors are connected in series,

Resistance of all four series connected resistors =4R

current through the circuit =E/4R

Then power dissipated in all the four resistors P_T=E×E/4R=20

P_T=E²/4R=20

Then
E²/R=80______(1)

When resistors are connected in parallel

Voltage across each resistor will be =E

Current through each resistor =E/R

Then, power consumed by each resistor P=V×I=E×E/R=E²/R

By substituting value of E²/R from equation (1) in the above relation, we get

P=E²/R=80 Watts

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