Question

in a metre-bridge experiment, 2 resistances
P and Q are connected in series in the
left gap. When the resistance in the right
gap is 50 ohm, the balance the points is at
the centre of the slide wire . if P and Q
are connected in paralled in the left gap.
the resistance in the
in the right gap has to
be changed to 12 ohm so as to obtain the
balance point at the same position.
Find P and Q.​

Answer 1

Given:

Two resistances P and Q has been placed in the left gap of metre Bridge experiment. When P and Q series combination the resistance in the right gap is 50 ohm and the balance point is at the centre of the slide wire. When P and Q add in parallel combination in the left gap, 12 ohm has to be given in the right gap for same position of balance.

To find:

Value of P and Q.

Calculation:

During series combination:

$\rm{ \dfrac{P + Q}{50} = \dfrac{50 \: cm}{50 \: cm} }$

$\rm{ = > \dfrac{P + Q}{50} = 1 }$

$\rm{ = > P + Q = 50 \: \: \: .......(1)}$

During parallel combination:

$\sf{ \dfrac{ (\frac{PQ}{P + Q}) }{12} = \dfrac{50 \: cm}{50 \: cm} }$

$\sf{ = > \dfrac{ (\frac{PQ}{P + Q}) }{12} = 1 }$

$\sf{ = > \dfrac{PQ}{P + Q} = 12 }$

$\sf{ = > \dfrac{PQ}{50} = 12 }$

$\sf{ = > PQ = 600 \: \: \: \: .......(2)}$

Putting value of Q in eq(1):

$\rm{ = > P + Q = 50 }$

$\rm{ = > P + \dfrac{600}{P} = 50 }$

$\rm{ = > {P}^{2} - 50P + 600 = 0 }$

$\rm{ = > (P - 30)(P + 20) = 0 }$

$\boxed{\sf{ \therefore \: P = 30 \: ohm}}$

$\sf{P + Q = 50}$

$\sf{ = > 30 + Q = 50}$

$\boxed{ \sf{ = > Q = 20 \: ohm}}$