Question

The equivalent resistance between A and B for the circuit shown in figure is

• A. $T_{1} > T_{2}$
• B. $T_{1} < T_{2}$
• C. $T_{1} = T_{2}$
• D. $T_{1} = \frac{1}{T_{2}}$

Answer 1

Answer: (A)

$T_{1} > T_{2}$

Answer 2

: For equivalent resistance between A and b. $5\Omega$and $8\Omega$resistances are connected in series. $R',$ their equivalent resistance is parallel to $6\Omega$

$\therefore R' = 5 + 8 = 13\Omega$

and, $\frac{1}{R''} = \frac{1}{13} + \frac{1}{6} = \frac{6 + 13}{78} = \frac{19}{78}$

$R'' = \frac{78}{19}$

Now $4\Omega,R''$and $5\Omega$resistances are connected in series equivalent resistance between A and B.

$\therefore R_{eq} = 4 + \frac{78}{19} + 5 = \frac{76 + 78 + 95}{19} = 13.1\Omega$