A cell of emf 10 V and internal resistance 3 Ω is connected to a uniform wire of length 500 cm and reistance 3 Ω. The drop of potential with length in the wire is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 10 mV/cm
Given:
EMF of cell $E = 10 \, V$
Internal resistance $r = 3 \, \Omega$
Resistance of wire $R = 3 \, \Omega$
Length of wire $L = 500 \, cm$
Total resistance $= R + r = 3 + 3 = 6 \, \Omega$
Current in the circuit: $I = \frac{E}{R+r} = \frac{10}{6} = \frac{5}{3} \, A$
Potential drop across the wire: $V_{wire} = I \times R = \frac{5}{3} \times 3 = 5 \, V$
Potential drop per unit length: $\frac{V_{wire}}{L} = \frac{5}{500} = 0.01 \, V/cm$
Final Answer: Potential drop in the wire $= 0.01 \, V/cm$