Advertisements
Advertisements
प्रश्न
Answer the following question.
What is the end error in a meter bridge? How is it overcome? The resistances in the two arms of the metre bridge are R = Ω and S respectively. When the resistance S is shunted with equal resistance, the new balance length found to be 1.5 l1, where l2 is the initial balancing length. calculate the value of s.
Advertisements
उत्तर
The shifting of zero of the scale at different points as well as the stray resistance gives rise to the end error in meter bridge wire. This error arises due to the non-uniformity of the meter wire End corrections can be estimated by including known resistances P1 and Q1 in the two ends and finding the null point.
We have
R = 5 Ω
According to wheat stone bridge principle:
`R/l_1 = S/(100 - l_1)`
`5/l_1 = S/(100 - l_1)` ....(1)
After shunting, connect the resistance in parallel
`S → S/2`
`5/(1.5 l_1) = S/(2(100 - 1.5 l_1))` ....(2)
Thus, equation (1) can be written as
`500 - 5l_1 = Sl_1` ....(3)
And equation 2 is
`10(100 - 1.5l_1) = 1.5 Sl_1` ....(4)
From equation (3) and (4)
`(500 - 5l_1)/l_1 = (1000 - 15l_1)/(1.5l_1)`
`750 - 7.5l_1 = 1000 - 15l_1`
`-250 = -7.5l_1`
`l_1 = 100/3`
S = `(500 - (5 xx 100)/3)/(100/3) = (500 - 500/3)/(100/3) = 1000/3 xx 3/100`
S = 10 Ω
संबंधित प्रश्न
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?
Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance?
A rectangular conductor LMNO is placed in a uniform magnetic field of 0.5 T. The field is directed perpendicular to the plane of the conductor. When the arm MN of length of 20 cm is moved towards left with a velocity of 10 ms−1, calculate the emf induced in the arm. Given the resistance of the arm to be 5 Ω (assuming that other arms are of negligible resistance) find the value of the current in the arm.
A cell of emf E and internal resistance r is connected to two external resistance R1 and R2 and a perfect ammeter. The current in the circuit is measured in four different situations:
(i) without any external resistance in the circuit
(ii) with resistance R1 only
(iii) with R1 and R2 in series combination
(iv) with R1 and R2 in parallel combination
The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in the order. Identify the currents corresponding to the four cases mentioned above.
Two non-ideal batteries are connected in series. Consider the following statements:-
(A) The equivalent emf is larger than either of the two emfs.
(B) The equivalent internal resistance is smaller than either of the two internal resistances.
A battery of emf 100 V and a resistor of resistance 10 kΩ are joined in series. This system is used as a source to supply current to an external resistance R. If R is not greater than 100 Ω, the current through it is constant up to two significant digits.
Find its value. This is the basic principle of a constant-current source.
Do the electrodes in an electrolytic cell have fixed polarity like a battery?
A conductor of length ‘l’ is rotated about one of its ends at a constant angular speed ‘ω’ in a plane perpendicular to a uniform magnetic field B. Plot graphs to show variations of the emf induced across the ends of the conductor with
- angular speed ω and
- length of the conductor l.
A straight line plot showing the terminal potential difference (V) of a cell as a function of current (I) drawn from it, is shown in the figure. The internal resistance of the cell would be then ______.
The internal resistance of a cell is the resistance of ______
