" Q.60.A resistance of "R Omega" draws cu "

A resistance of R Omega draws current from a potentiometer. The potentiometer has a total resistance R _(0) Omega. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer.

A resistance of R Omega draws current from a potentiometer. The potentiometer has a total resistance R_(0) Omega . A voltage V is supplied to the potentiometer. Dervie can expression for the voltage across R when the sliding contact is in the middle of the potentiometer.

With a resistance R connected in series with a galvanometer of resistance 100 Omega , it acts as a voltmeter of range 0-V. To double the range a resistance of 1000 Omega is to be connected in series with R. Then the value of R is ( Omega )

The resistance of a heater coil is 110 Omega . A resistance R is connected in parallel with it and the combination is joined in series with a resistance of 11 Omega to a 220 V main line. The heater operates with a power of 110 W . The value of R in Omega is

A transistor connected in common emitter configuration has input resistance R_("in")=2 K Omega and load resistance of 5 K Omega . If beta=60 and an input signal 12 mV is applied , calculate the resistance gain.

Two resistors of resistance R_1 = 2 Omega and R_2= 1 Omega are connected in parallel with a current source of 3 A. Draw the arrangement and deduce the current I_1" in "R_1 and I_2" in "R_2.