In the figure, each resistance is of 10 ohms. The equivalent resistance between P and R is ?

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1 Answer
Apr 30, 2018

R_"eq final" = 27.5Omega

Explanation:

Let's have symbols for all nodes. From P, the first node to the right shall be P_1, the next to the right shall be P_2, and the 3rd to the right is already known as Q. From R, the first node to the right shall be R_1, the next to the right shall be R_2, and the 3rd to the right is already known as S.

Since there is no connection from Q toward the right, the path thru the resistor between P_2 and Q can have no current thru it. The same can be said about the path thru the resistor between R_2 and S. Therefore we can ignore the 2 resistors farthest to the right.

Let's evaluate smaller portions of the circuit and calculate their equivalent resistance. And we will work our way thru the circuit using that equivalent resistance to form another portion of the circuit to form another equivalent resistance incorporating more of the original circuit. Call the first equivalent resistance determined R_"eq1", the next R_"eq2" and so on.

Now, notice the 3 resistors in the path P_1, P_2, R_1, and R_2. Those 3 resistors are in series, so the equivalent resistance, R_"eq1", of those 3 is given by

R_"eq1" = 10Omega+10Omega+10Omega = 30 Omega

Those 30 Omegas are in parallel with the 10 Omega directly between P_1 and R_1. The equivalent resistance, R_"eq2", of that parallel combination is given by

R_"eq2" = (30 Omega * 10 Omega)/(30 Omega + 10 Omega) = 7.5 Omega

The resistor between P and P_1, the equivalent resistance, R_"eq2"; and the resistor betweenR and R_1, form a series combination. The equivalent resistance, R_"eq final", of that series combination will be the final answer to the question. It is given by

R_"eq final" = 10Omega+7.5Omega+10Omega = 27.5Omega

I hope this helps,
Steve