Q20 of 30 Page 128

Determine the equivalent resistance of networks shown in Fig. 3.31.

(a)



It can be observed from the given circuit that in the first small loop, two resistors


of resistance 1 Ω each are connected in series.


Hence, their equivalent resistance = (1 + 1) = 2 Ω


It can also be observed that two resistors of resistance 2 Ω each are connected in series.


Hence, their equivalent resistance = (2 + 2) = 4 Ω.


Now the given circuit diagram can be explained as


Now draw 4 loops each having 2 resistors of 2 Ω and 4 Ω connected in parallel.


Hence Equivalent Resistance Re of each loop is


⇒ Re = 4/3 Ω


Since all 4 loops will be connected in series and each loop will have 4/3 Ω resistance


Hence, equivalent resistance of the given circuit is 4 × 4/3 = 16/3 Ω .


(b)


From the given circuit that five resistors of resistance R each are connected in series.


Hence, equivalent resistance of the circuit = R + R + R + R + R = 5R


More from this chapter

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20

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?

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Given the resistances of 1Ω, 2Ω, 3Ω, 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?

 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.

 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 e 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 kW 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?