A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Emf of the battery, E = 10 V
Internal resistance of the battery, r = 3 Ω
Current in the circuit, I = 0.5 A
Resistance of the resistor = R
The relation for current using Ohm's law is,
Terminal voltage of the resistor = V
According to Ohm's law,
V = IR
= 0.5 x 17
= 8.5 V
Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is
8.5 V.
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An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.
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(b) Is there a transfer of mass from wool to polythene?
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
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A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.
A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere?
A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
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The carrier wave is given by
(i) Sketch the amplitude modulated waveform
(ii) What is the modulation index?
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Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf Ωµ and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
(a) What is the value Ωµ ?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
(f ) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?
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(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.
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(i) Nature of the source.
(ii) Direction of propagation.
(iii) Motion of the source and/or observer.
(iv) Wave length.
(v) Intensity of the wave.
On which of these factors, if any, does
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Thank you so much sir it's really helpful
Thanks sir , good explanation
It's very helpful for physics students