Question
$2 \mathrm{~A}$ current is obtained when a $2 \Omega$ resistor is connected with battery having $r \Omega$ as internal resistance $0.5 \mathrm{~A}$ current is obtained if the above battery is connected to $9 \Omega$ resistor. Calculate the internal resistance of the battery.(A) $0.5 \Omega$(B) $(1 / 3) \Omega$(C) $(1 / 4) \Omega$(D) $1 \Omega$
Step 1
Step 1: We know that the electromotive force (emf) of a battery is given by the formula: \[E = I(R + r)\] where \(I\) is the current, \(R\) is the resistance connected to the battery, and \(r\) is the internal resistance of the battery. Show more…
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A 5 V battery with internal resistance $2 \Omega$ and $2 v$ battery with internal resistance $1 \Omega$ are connected to $10 \Omega$ resistor as shown in fig the current in $10 \Omega$ resistor is.... (A) $0.27 \mathrm{~A}, \mathrm{P}_{1}$ to $\mathrm{P}_{2}$ (B) $0.27 \mathrm{~A}, \mathrm{P}_{2}$ to $\mathrm{P}_{1}$ (C) $0.03 \mathrm{~A}, \mathrm{P}_{1}$ to $\mathrm{P}_{2}$ (D) $0.03 \mathrm{~A}, \mathrm{P}_{2}$ to $\mathrm{P}_{1}$
A $5 \mathrm{~V}$ battery with internal resistance $2 \Omega$ and $2 \mathrm{~V}$ battery with internal resistance $1 \Omega$ are connected to a $10 \Omega$ resistor as shown in figure. The current in $10 \Omega$ resistor is (a) $0.27 \mathrm{~A}, P_{1}$ to $P_{2}$ (b) $0.27 \mathrm{~A}, P_{2}$ to $P_{1}$ (c) $0.03 \mathrm{~A}, P_{1}$ to $P_{2}$ (d) $0.03 \mathrm{~A}, P_{2}$ to $P_{1}$
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Round 2
A $5 \mathrm{~V}$ battery with internal resistance $2 \Omega$ and a $2 \mathrm{~V}$ battery with internal resistance $1 \Omega$ are connected to a $10 \Omega$ resistor as shown in the figure. The current in the $10 \Omega$ resistor is (a) $0.27 \mathrm{~A}, P_{2}$ to $P_{1}$ (b) $0.03 \mathrm{~A}, P_{1}$ to $P_{2}$ (c) $0.03 \mathrm{~A}, P_{2}$ to $P_{1}$ (d) $0.27 \mathrm{~A}, P_{1}$ to $P_{2}$
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