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3.20 b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (iii) 6 Ω

Attach them in series

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.

Given,  the balance point of cell in open circuit =  value of external resistance added =  new balance point =  let the internal resistance of the cell be . Now as we know,  in a potentiometer, Hence, the internal resistance of the cell will be 1.68

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 upto 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 ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(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?

No, the circuit would not work properly for very low order of Voltage because, the balance point would be near point A and there will be more percentage error in measuring it. If we add series resistance with wire AB. It will increase the potential difference of wire AB which will lead to a decrease in percentage error.

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 upto 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 ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(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?

No, the method would not have worked if the driver cell of the potentiometer had an emf of 1.0V instead of 2, because when emf of the driving point is less than the other cell, their won't be any balance point in the wire

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 upto 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 ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(c) Is the balance point affected by this high resistance?

No, the Balance point is not affected by high resistance. High resistance limits the current to galvanometer wire. The balance point is obtained by moving the joe key on the potentiometer wire and current through potentiometer wire is constant. The balance point is the point when the current through galvanometer becomes zero. The only duty of high resistance is to supply limited constant...

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 upto 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 ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(b) What purpose does the high resistance of 600 k$\Omega$ have?

If a sufficiently high current passes through galvanometer then it can get damaged. so we limit the current by adding a high resistance of 600 k.

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 ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value of $\epsilon$

Given maintained  constant emf of standard cell  = 1.02V, balanced point of this cell = 67.3cm Now when the standard cell is replaced by another cell with emf = . balanced point for this cell = 82.3cm Now as we know the relation  Hence emf of another cell is 1.247V.

3.21 Determine the current drawn from a 12V supply with internal resistance 0.5Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.

First, let us find the equivalent of the infinite network, let equivalent resistance = R' Here from the figure, We can consider  the box as a resistance of R' Now, we can write, Equivalent resistance = R' ' =[( R')Parallel with (1)] + 1 + 1        Since resistance can never be negative we accept   , We have calculated the equivalent resistance of infinite network, NOW Total Equivalent...

3.20 (c) Determine the equivalent resistance of networks shown in Fig. 3.31.

(b)

It can be seen that all 5 resistor are in series, so Equivalent Resistance = R + R + R + R + R = 5R Hence equivalent resistance is 5R.

3.20 (c) Determine the equivalent resistance of networks shown in
Fig. 3.31.

(a)

It can be seen that in every small loop resistor 1 ohm is in series with another 1 ohm resistor and two 2 ohms are also in series and we have 4 loops, so equivalent resistance of one loop is equal to the parallel combination of 2 ohms and 4 ohm that is  now we have 4 such loops in series so,  Hence equivalent resistance of the circuit is 16/3 ohm.

3.20 (b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them
to get an equivalent resistance of (iv)$\frac{6}{11}\Omega$

Connect all three resistors in parallel. Equivalent Resistance is  R =  1/(1/1 + 1/2 + 1/3) = (1x 2 x 3)/(6 + 3 + 2) R = 6/11 Ω

3.20 (b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine the to get an equivalent resistance of (ii) $\frac{11}{5}\Omega$

Connect 2 Ω and 3 Ω resistor in parallel and 1 Ω resistor in series to it Equivalent Resistance R = {1/(1/2 + 1/3)} + 1 = 6/5 + 1 R = 11/5 Ω

3.20 (b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them
to get an equivalent resistance of (i) $\frac{11}{3}\Omega$

We Have, equivalent resistance = 11/3 Let's break this algebraically so that we can represent it in terms of 1, 2 and 3 this expression is expressed in terms of 1, 2 and 3. and hence we can make a circuit which consist only of 1 ohm, 2 ohms and 3 ohms and whose equivalent resistance is 11/3. that is :

20 a (iii)What is the ratio of the maximum to minimum resistance?

The ratio is nR/(R/n) =

3.20 (a) Given n resistors each of resistance R, how will you combine them to get the (ii) minimum effective resistance?

To get minimum effective resistance, combine them in parallel. The effective resistance will be R/n.

3.20 (a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum effective resistance?

To get maximum effective resistance, combine them in series. The effective resistance will be nR.

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 ($10^{22} / 10^{23}$ ).

(a) Alloys of metals 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 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 .

(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

A very high internal resistance is required for a high tension supply to limit the current drawn for safety purposes.