A Wheatstone bridge arrangement is shown as below:
Using Kirchoff’s second law to the loop ABDA, we get
Applying Kirchoff’s law to loop BCDB, we get
When the bridge is balanced,
Then, the equations can be written as,
On dividing equation (1) by (2), we get
, which is the balanced condition of a Wheatstone bridge.
Name the parts of the electromagnetic spectrum which is
(a) suitable for radar systems used in aircraft navigation.
(b) used to treat muscular strain.
(c) used as a diagnostic tool in medicine.Write in brief, how these waves can be produced.
An inductor L of inductance XL is connected in series with a bulb B and an ac source.How would brightness of the bulb change when (i) number of turn in the inductor is reduced, (ii) an iron rod is inserted in the inductor and (iii) a capacitor of reactance XC = XL is inserted in series in the circuit. Justify your answer in each case.
i) Net resistance in the circuit is given by,
Inductance is given by,
As number of turns decreases, L decreases.
Inductance is given by,
Therefore, will also decrease thereby, reducing the net resistance in the circuit. Thus, current increases and brightness of the bulb is increased.
ii) When a soft iron rod is inserted in the circuit, L increases. Therefore, inductive reactance also increases. Net resistance increases and flow of current in the circuit decreases. Thus, the brightness of the bulb will decrease.
iii) When a capacitor of reactance XL = XC is connected in series with the circuit net resistance becomes,
; where R is the resistance of the bulb.
Here, we will have Z= R which is the condition of resonance.
At resonance, maximum current will flow through the circuit. Therefore, the brightness of the bulb will increase.
Here, a concave lens is placed in a medium of refractive index greater than the refractive index of the material of the lens.
So, the given lens will behave as a diverging lens.
A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus (i) R and (ii) the current I.It is found that when R = 4 , the current is 1 A and when R is increased to 9 , the current reduces to 0.5 A. Find the values of the emf E and internal resistance r.
Plot for V vs. R
Plot for V vs. I
Current in the circuit is given by,
The resistance offered by the capacitor when connected to an electrical circuit is known as capacitive reactance.
It is given by,
is the angular frequency of the source, and
C, is the capacitance of the capacitor.
SI Unit is ohm.
Two capacitors of unknown capacitances C1 and C2 are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of C1 and C2. Also, calculate the charge on each capacitor in parallel combination.
When the capacitors are connected in parallel,
Energy stored in capacitors, EP =
When the capacitors are connected in series,
Putting this in equation (2), we get
C2 = 50 – 11.2 = 38.2 μF
The charge on capacitor is given by, Q = CV
Charge on capacitor 1, C1 =
Charge on capacitor 2, C2 =
State the principle of working of a galvanometer.A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a resistance R1 in series with the coil. If a resistance R2 is connected in series with it, then it can measure upto V/2 volts. Find the resistance, in terms of R1 and R2, required to be connected to convert it into a voltmeter that can read upto 2 V. Also find the resistance G of the galvanometer in terms of R1 and R2.
When a current carrying loop or coil is placed in the uniform magnetic field, moving coil galvanometer experiences a torque.
A high resistance is connected in series with the galvanometer to convert it into a voltmeter.
The value of resistance is given by,
V is the potential difference across the terminals of the voltmeter.
is current through galvanometer and
G is the resistance of the galvanometer.
When resistance R1 is connected in series with the galvanometer,
When resistance R2 is connected in series with the galvanometer,
From equations (1) and (2), we get
Resistance R3 required to convert galvanometer into voltmeter of range 0 to 2V is given by,
Therefore, R1 - 2R2 is the galvanometer resistance in terms of R1 and R2
Graph showing the variation of current versus voltage for a material GaAs is shown in the figure. Identify the region of
i) negative resistanceii) where Ohm’s law is obeyed.
i) The region of negative resistance is DE because, the slope is negative for this part of curve.
ii) BC is the part of the curve where Ohm’s law is obeyed because here, current is varying linearly with with the voltage. This gives us direct proportionality between current and voltage.
Given, cube encloses an electric dipole.
So, the total charge enclosed by the cube is zero.
That is, Q= 0.
Now, using the formula for electric flux as per the Gauss’s law of electrostatics, we have