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Question:

A cell of emf E and internal resistance r is connected to two external resistances R1 and R2 and a perfect ammeter. The current in the circuit is measured in four different situations: (i) without any external resistance in the circuit (ii) with resistance R1 only (iii) with R1 and R2 in series combination (iv) with R1 and R2 in parallel combination. The currents measured in the four cases are 0.42A, 1.05A, 1.4A and 4.2A, but not necessarily in the order. Identify the currents corresponding to the four cases mentioned above.

Solution:

Using Ohm's law, V=IR
for (i) : V=E, I=I1, R=r so I1=E/r
for (ii): V=E, I=I2, R=r+R1 so I2=E/(r+R1)
for (iii): V=E, I=I3, R=r+(R1+R2) so I3=E/(r+R1+R2)
for (iv): V=E, I=I4, R=r+(R1R2/(R1+R2)) so I4=E/(r+(R1R2/(R1+R2)))
Thus ,I1>I4>I2>I3 (The equivalent resistance for parallel combination is less than of each resistance )
Hence, 4.2 A is for (i), 1.4 A for (iv), 1.05 A for (ii) and 0.42 A for (iii).