Such problems involving equivalent resistance and capacitance have been regularly set in previous IITJEE and AIEEE exams in some form or the other. So practice of such problems is a must before appearing in similar tests. Moreover, attempts to solve such problems provide a good understanding of electrical circuits and help in solving other circuit problems too.

Find the equivalent capacitance across the principal diagonal AE of the mesh cube shown in fig. 1 which has a resistance ‘R’ in each side.

There are many techniques to solve this problem but the simplest one will be described here.

Let us extend the terminal A and E and connect them to a voltage source ‘V’ and also assume that a current 6A passes in the main circuit as shown in fig 2.

Now, at point A the main current of 6 A divides into 3 branches and due to symmetry we can understand that each branch will carry a current of 2A (Branches AE, AB and AD).

Now the branch AB supplies current to BF and BC. Again by symmetry we can understand that BF and BC will carry currents of 1A each. Branches BF, AC, EF, EH,DH and DC are all 1A branches.

The 1A currents in these branches combine to re-create 2A currents in branches HG, FG and CG which in turn combine to re-create the main 6A current.

Now, let us apply KVL to the loop ADCG-V, we get,

V=2R+R+2R

ð V=5R

ð Since, V=6 R EQ. where R Eq= Total resistance of the cube mesh, we have,

6 R Eq=5R

Therefore, R Eq=(5/6)R

Ans= 5R/6

Find the equivalent capacitance across the points ‘P’ and ‘Q’ in the network of the adjacent figure.

Such problems which involve capacitances or resistances difficult to understand whether they are in parallel or series, can be solved by a method of writing terminal voltages and then grouping the capacitances with similar voltage differences together.

In figure 2, the end voltages have been assumed as v1, v2, v3. Mark that ends of capacitors joined by a conductor will have same voltage.

If we group together capacitances with same end voltages we get the following :

End voltages v1 and v2: 3 number of capacitors= 3 capacitors in parallel=3C

End voltages v2 and v3: 2 number of capacitors=2 capacitors in parallel=2C

End voltages v2 and v4: 1 number of capacitor= C

Now, the diagram can be redrawn as in figure 3.

Since we have to find the equivalent capacitance between P and Q, we take the path PSQ and neglect the part SR.

This means 3C and C in series.

Therefore, Equivalent Capacitance, C_eq=(3CXC)/(3C+C)=3C/4

Ans= 3C/4

I shall advise you all to first attempt to solve the questions yourself. Even if you find them difficult, try hard. Such attempts will help you in the long run.

If you want more questions of similar type with answers, i can provide. Please ask in the comments section if you want.

If you have any doubts, you can ask in the comments section. I will reply.