Three 10Ω, 2 W resistors are connected as in Fig. The maximum possible voltage between points A and B without exceeding the power dissipation limits of any of the resistors is:
Two light bulbs shown in the circuit have ratings A(24 V, 24 W) and B (24 V and 36 W) as shown. When the switch is closed.
The three resistances A, B and C have values 3R, 6R and R respectively. When some potential difference is applied across the network, the thermal powers dissipated by A, B and C are in the ratio :
An electric bulb rated for 500 watts at 100 volts is used in a circuit having a 200-volt supply. The resistance R that must be put in series with the bulb, so that the bulb draws 500 watts is :
To find the resistance
that must be placed in series with the bulb, let's analyze the problem step by step.
) = 500 W,
) = 100 V,
) = 200 V.
The resistance of the bulb (
) can be calculated using the formula:
Substitute the values:
The bulb is rated to draw 500 W at 100 V. Thus, the current through the bulb is:
Substitute the values:
The total supply voltage is 200 V, and the bulb operates at 100 V. Therefore, the voltage drop across the series resistor
is:
Substitute the values:
Using Ohm's law, the resistance of the series resistor is:
Substitute the values:
The resistance that must be placed in series with the bulb is:
An ammeter is to be constructed which can read currents upto 2.0 A. If the coil has a resistance of 25 Ω and takes 1 mA for full-scale deflection, what should be the resistance of the shunt used?
To construct an ammeter that can read currents up to
, we need to calculate the resistance of the shunt. Here's the step-by-step calculation:
for full-scale deflection.
, the current through the shunt is:
. Using Ohm's law, the voltage across the coil is:
Thus, the resistance of the shunt is:
On the basis of electrical conductivity, which one of the following material has the smallest resistivity?
Silver is a metal which has very high electrical conductivity and hence possesses the smallest resistivity compared to semiconductors (Silicon, Germanium) and insulators (Glass).
A certain wire \(A\) has resistance \(81 \Omega\). The resistance of another wire \(B\) of same material and equal length but of diameter thrice the diameter of \(A\) will be
Resistance is inversely proportional to the square of the diameter: \(R = \frac{1}{d^2}\). Since the diameter is tripled, the resistance becomes \(\frac{R}{9} = \frac{81}{9} = 9 \Omega\).
A copper wire of radius 1 mm contains \(10^{22}\) free electrons per cubic metre. The drift velocity for free electrons when 10 A current flows through the wire will be (Given, charge on electron = \(1.6 \times 10^{-19}\text{ C}\))
Using the relation \(I = n e A v_d ⇒ v_d = \frac{I}{n e \pi r^2}\). Substituting the given values: \(v_d = \frac{10}{10^{22} \times 1.6 \times 10^{-19} \times \pi \times (10^{-3})^2} = \frac{6.25}{\pi} \times 10^3\text{ m s}^{-1}\).
If a copper wire is stretched to make it 0.1% longer, the percentage increase in resistance will be:
Resistance of a stretched wire of constant volume is proportional to the square of its length: \( R \propto l^2 \). Thus, \( \frac{\Delta R}{R}\text{ (%)} = 2 \frac{\Delta l}{l}\text{ (%)} = 2 \times 0.1\% = 0.2\% \).
A galvanometer has a coil of resistance \( 100\ \Omega \) and gives full scale deflection for 20 mA current. If it is to work as a voltmeter of 10 volt range, the resistance required to be added will be:
To convert a galvanometer to a voltmeter, a high series resistance \( R \) is added: \( R = \frac{V}{I_g} - G \). Substituting values, \( R = \frac{10}{20 \times 10^{-3}} - 100 = 500 - 100 = 400\ \Omega \).