A battery with negligible internal resistance is connected to a resistor. The po

Question

A battery with negligible internal resistance is connected to a resistor. The power produced in the battery and the power dissipated in the resistor are both rho_1 Eight resistors of the same kind are added, so the circuit consists of a battery and nine resistors in series. In term of P_1, now how much power is dissipated in the first resistors? rho_ resistors = P_1/2 In terms of P_1, now how much power is produced in the battery? rho_ resistors = P_1/9 The circuit is rearranged so that the resistors are in parallel rather than in series. In terms of rho_1, now how much power is produced in the battery? rho_ resistor = 9P_1

Explanation / Answer

When one resistor of resistance R to a battery of emf E then current in the resistance i = E/R

Power in resistor P1 = i 2 R

= (E/R) 2 R

= E 2 / R (1)

When nine resistors of resistance R are connected in series to a battery of emf E then current in the resistance i = ?

Equivalent resistance Req = 9R

So, current in the circuit , i ' = E/Req

= E/9R

Power in first resistor P ' = i ' 2 R

= (E/9R) 2 R

= E 2 / 81R

= P1/81 Since from equation(1)

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