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Two cells of voltage 10 V and 2 V, and internal resistances 10 Ω and 5 Ω respectively, are connected in parallel with the positive end of 10 V battery connected to negative pole of 2 V battery (figure). Find the effective voltage and effective resistance of the combination.
Two cells of voltage 10 V and 2 V, and internal resistances 10 Ω and 5 Ω respectively, are connected in parallel with the positive end of 10 V battery connected to negative pole of 2 V battery (figure). Find the effective voltage and effective resistance of the combination.
This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation- applying kirchhoff's junction rule I1 = I+I2
Applying kirchhoff's rule in outer loop containing 10V cell
10= IR+10I1……………. (1)
Applying kirchhoff's rule in outer loop containing 2V cell
2= 5I2-RI= 5 (I1-I)-RI
4= 10I1-10
Upvote 
(a) Consider circuit in Fig 3.10. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity?
(b) Electrons give up energy at the rate of RI2 per second to the thermal energy. What time scale would one associate with energy in problem (a)? n = no of electron/volume = 1029/m3, length of circuit = 10 cm, cross-section = A = (1mm)2
(a) Consider circuit in Fig 3.10. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity?
(b) Electrons give up energy at the rate of RI2 per second to the thermal energy. What time scale would one associate with energy in problem (a)? n = no of electron/volume = 1029/m3, length of circuit = 10 cm, cross-section = A = (1mm)2
This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation- according to ohm's law V= IR
I= 6/6 = 1A
I= AneVd or Vd= i/neA
Upvote In an experiment with a potentiometer, VB = 10 V. R is adjusted to be 50 Ω (figure). A student wanting to measure voltage E1of a battery (approx. 8 V) finds no null point possible. He then diminishes R to 10 Ω and is able to locate the null point on the-last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.
In an experiment with a potentiometer, VB = 10 V. R is adjusted to be 50 Ω (figure). A student wanting to measure voltage E1of a battery (approx. 8 V) finds no null point possible. He then diminishes R to 10 Ω and is able to locate the null point on the-last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.
This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation – let R' be the resistance of potentiometer wire.
Effective resistance of potentiomter and variable resistor r=50ohm is 500+R'
Effective voltage across potentiometer = 10V
The current through main circuit I= =
Potential
Upvote A room AC runs for 5 a day at a voltage of 220 V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions?
A room AC runs for 5 a day at a voltage of 220 V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions?
This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation – power consumption in a day i.e in 5 = 10 units
Power consumption per hour = 2 units
Power consumption = 2 units =2KW= 2000J/s
Also power =V I
2000W= 220V l or l= 9A approx.
R=
Power consumption in first curre
Upvote A room AC runs for 5 a day at a voltage of 220 V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions?
A room AC runs for 5 a day at a voltage of 220 V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions?
This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation – power consumption in a day i.e in 5 = 10 units
Power consumption per hour = 2 units
Power consumption = 2 units =2KW= 2000J/s
Also power =V I
2000W= 220V l or l= 9A approx.
R=
Power consumption in first curre
Upvote 3.23 Figure 3.34 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
3.23 Figure 3.34 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
3.23 Internal resistance of the cell = r
Balance point of the cell in open circuit, = 76.3 cm
An external resistance R is connected to the circuit with R = 9.5 Ω
New balance point of the circuit, = 64.8 cm
Current flowing through circuit = I
The relation connecting resistance and em
Upvote 3.22 Figure 3.33 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 and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
(a) What is the value e ?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?
(e) 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?
3.22 Figure 3.33 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 and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
(a) What is the value e ?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?
(e) 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?
3.22 Constant emf of the standard cell, = 1.02 V
Balance point on the wire, = 67.3 cm
A cell of unknown emf, , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm.
The relation of connected emf and balance point is, =
Hence,
Upvote 3.21 Determine the current drawn from a 12V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.
3.21 Determine the current drawn from a 12V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.
3.21 Let the equivalent resistance of the given circuit be R’. The equivalent resistance of an infinite network is given by
R’ = 2 + or R’ = or + R’ = 2R’ + 2 + R’
R’ = = 1
Since R’ cannot be negative, hence R’ = 1+ = 2.73 Ω
Internal resistanc
Upvote 3.20 (a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?
(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?
(c) Determine the equivalent resistance of networks shown in Fig. 3.31.
3.20 (a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?
(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?
(c) Determine the equivalent resistance of networks shown in Fig. 3.31.
3.20 Total number of resistors = n
Resistance of each resistor = R
When the resistors are connected in series, effective resistance is maximum. = nR
When n resistors are connected in parallel, the effective resistance, is minimum. = . The ratio of
Upvote Q.3.19 Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (1022/1023).
Q.3.19 Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (1022/1023).
3.19 (a) Alloys of metal usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have much lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of increase of temperature.
(d)
Upvote