166. A 35 V d.c. supply is connected across a combined resistance of 600 ohms and an unknown resistance of R ohms in series. A voltmeter having a resistance of 1.2 kW is connected across 600 ohm resistor and reads 5 V. The resistance R will be
     

1. 1. 120 ohms

   
   
   

   2. 500 ohms

   
   
   

   3. 1.7 k ohms

   
   
   

   4. 2.4 k ohms

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Parallel sum of 600 ohm and 1200 ohm
   

   =	600 × 1200	= 400 ohm
   	1800

   
   

   ∴	R	=	400
   	35.5		5

   
   or R = 2400 ohms = 2.4 k ohms

   Correct Option: D

   Parallel sum of 600 ohm and 1200 ohm
   

   =	600 × 1200	= 400 ohm
   	1800

   
   

   ∴	R	=	400
   	35.5		5

   
   or R = 2400 ohms = 2.4 k ohms

---

167. A thermistor exhibits

1. 1. only a negative change of resistance with increase in temperature

   
   
   

   2. only a positive change of resistance with increases in temperature

   
   
   

   3. can exhibit either a negative or positive change of resistance with increase of temperature depending upon the type of material

   
   
   

   4. none of these

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

---

---

168. A wattmeter is reading back-wards in an experiment. Upscale reading can be obtained by reversing
     

1. 1. pressure coil connection only

   
   
   

   2. current coil connection only

   
   
   

   3. both pressure and current coil connection only.

   
   
   

   4. either pressure or current connection

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

---

169. A metal strain gauge factor of two. Its nominal resistance is 120 ohms. If it undergoes a strain of 10^–5, the value of change of resistance in response to the strain is

1. 1. 240 ohms

   
   
   

   2. 2 × 10⁵ohm

   
   
   

   3. 2.4 × 10^–3 ohm

   
   
   

   4. 1.2 × 10^–3 ohm

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Gauge factor =	dR/R
   	ξa

   
   

   ∴	dR	= 2 × 10-5
   	R

   
   or dR = 2 × 120 × 10^–5 = 2.4 × 10^–3

   Correct Option: C

   Gauge factor =	dR/R
   	ξa

   
   

   ∴	dR	= 2 × 10-5
   	R

   
   or dR = 2 × 120 × 10^–5 = 2.4 × 10^–3

---

---

170. Some wire-wound resistors have bifilar winding. This type of winding is used to

1. 1. increase the thermal stability

   
   
   

   2. reduce the tolerance

   
   
   

   3. reduce the inductance of winding

   
   
   

   4. double the power rating of the resistance

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

---