Tag: Kirchoff’s laws

Power dissipated in resistance networks

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JEE Advanced 2008 Paper 1, Question 25

The figure below shows three resistor configurations R1, R2 and R3 connected to 3 \mathrm{~V} battery. If the power dissipated by the configuration R1, R2 and R3 is P1, P2 and P3 respectively, then

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  1. P1 > P2 > P3
  2. P1 > P3 > P2
  3. P2 > P1 > P3
  4. P3 > P2 > P1

Solution

The power dissipated by a resistance network is

(1)   \begin{equation*} P = \frac{V^2}{R} . \end{equation*}

The potential applied to all three networks arre the same, so we just need to find the resistances. Of circuits R2 and R3, the latter has a greater resistance since

(2)   \begin{equation*} % R_3 = 1 \, \Omega + \frac{2 \, \Omega \cdot 2 \, \Omega}{2 \, \Omega + 2 \, \Omega} = 2 \, \Omega , R_3 = 1 \, \Omega + \left( \frac{1}{2 \Omega} + \frac{1}{2 \Omega} \right)^{-1} = 2 \, \Omega , \end{equation*}

whereas

(3)   \begin{equation*} % R_2 = \frac{1 \, \Omega \cdot 1 \, \Omega}{1 \, \Omega + 1 \, \Omega} = 0.5 \, \Omega . R_2 = \left( \frac{1}{1 \Omega} + \frac{1}{2 \Omega} + \frac{1}{2 \Omega} \right)^{-1} = 0.5 \, \Omega. \end{equation*}

The first circuit can be redrawn in the following way to make the comparison easier.…

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Resistor network with mirror symmetry

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JEE Advanced 2012 Paper 1, Question 12

For the resistance network shown in the figure, choose the correct option(s).

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  1. The current through PQ is zero.
  2. I_1 = 3 A.
  3. The potential at S is less than that at Q.
  4. I_2 = 2 A.

Solution

We can speed up the solution by noticing the mirror symmetry of the circuit. Let the current I_1 split into two parts I_a and I_b at the node X (see figure). Similarly, let the currents going into node Y be I_a' and I_b'. If we were to reverse the polarity of the 12 \, {\rm V} battery connected to the circuit, all currents would simply reverse direction, but their magnitudes would remain the …

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