Question

what is the effective resistance between P and Q in the given circuit​

Answer 1

Answer:

All the resistance less wires that have been attached in between the three pairs of 20,30 and 40,60 resistances can be removed due to the formation of a balanced Wheatstone Bridge.

So,

The problem is now simple enough to calculate the parallel resistance of resistances 10,50,100 and 50.

So,

1/R = 1/10 + 1/50 + 1/100 + 1/50 = 10+2+2+1/100 = 15/100

So,

Equivalent resistance across the points P and Q is 100/15 ohms = 20/3 = 6.67 Ohms

Answer 2

Answer:

$\large\boxed{\sf{6.67\;\Omega}}$

Explanation:

Observing the figure, we get to know that,

$\dfrac{20}{30} = \dfrac{40}{60} = \dfrac{2}{3}$

Also, there is no resistance between these pairs.

.°. It's a Wheatstone bridge.

Now, we have simply four resistors in parallel.

All the respective resistors are sum of two resistors in series.

1st resistance = 6+4 = 10 Ω

2nd resistance = 20+30 = 50 Ω

3rd resistance = 40+60 = 100 Ω

4th resistance = 20+30 = 50 Ω

Now, We know that, equivalent resistance in parallel combination is given by is given by,

$\large\boxed{\red{\dfrac{1}{{R}_{eq}}</p><p>= \dfrac{1}{{R}_{1}} + \dfrac{1}{{R}_{2}} + \dfrac{1}{{R}_{2}} + ..........}}$

Thus, for the given conditon, we have,

$= > \frac{1}{{R}_{eq}} = \frac{1}{10} + \frac{1}{50} + \frac{1}{100} + \frac{1}{50} \\ \\ = > \frac{1}{{R}_{eq}} = \frac{10 + 2 + 1 + 2}{100} \\ \\ = > \frac{1}{{R}_{eq}} = \frac{15}{100} \\ \\ = > {R}_{eq} = \frac{100}{15} \\ \\ = > {R}_{eq} = 6.67$

Hence, effective resistance between P and Q in the given circuit is 6.67 Ω