Q.
What is the equivalent resistance between the points $A$ and $B$ as in the picture below?
NTA AbhyasNTA Abhyas 2020
Solution:
In the given question condition of wheatstone bridge concept is balanced we can draw the circuit according to the concept
Again we can draw it by following diagram
It this satisfy the relation of wheatstone bidge $\frac{P}{Q}=\frac{R}{S}$ then unknown resistance 10Ω is negleted hence no current will be drawn from 10Ω resistance
where P = 6Ω, Q = 3Ω, R = 8Ω and S = 4Ω
Again we can draw the circuit
And we can calculate in this figure 6Ω and 3Ω are in series combination i.e 6Ω + 3Ω = 9Ω
similarly 8Ω and 4Ω are in series i.e 8Ω + 4Ω = 12Ω
And now 12Ω and 9Ω are in parallel combination hence we can calculate $\frac{12 Ω \times 9 Ω}{12 Ω + 9 Ω}$
(Because we calculate by parallel combination fomula $\frac{R_{1} \times R_{2}}{R_{1} + R_{2}}$ )
And further 7Ω resistance are in series combination hence we calculate by
$\frac{12 Ω \times 9 Ω}{12 Ω + 9 Ω}+7Ω$ = $\frac{85}{7}Ω$
Again we can draw it by following diagram
where P = 6Ω, Q = 3Ω, R = 8Ω and S = 4Ω
Again we can draw the circuit
And we can calculate in this figure 6Ω and 3Ω are in series combination i.e 6Ω + 3Ω = 9Ω
similarly 8Ω and 4Ω are in series i.e 8Ω + 4Ω = 12Ω
And now 12Ω and 9Ω are in parallel combination hence we can calculate $\frac{12 Ω \times 9 Ω}{12 Ω + 9 Ω}$
(Because we calculate by parallel combination fomula $\frac{R_{1} \times R_{2}}{R_{1} + R_{2}}$ )
And further 7Ω resistance are in series combination hence we calculate by
$\frac{12 Ω \times 9 Ω}{12 Ω + 9 Ω}+7Ω$ = $\frac{85}{7}Ω$