Old pagination

- A 1 kΩ, 1 W resistor can safely pass a current of
The wattage rating of the resistor W = I
^{2}.RHere W = 1 W, R = 1 kΩ = 1000 Ω∴ I = √(1/1000) = 0.0316 A = 31.6 mA > 30 mA. - Two resistors are connected in parallel across a battery of 2 V and a current flow through the combine resistors is 2 A. It one of the resistors is disconnected, the current will become 1.5 A, then what will be the resistance of that disconnected resistors?
Total current is 2A and after disconnection of one, resistors, the current drawn from the battery, is 1.5A. That means the disconnected resistors was sharing 0.5A of currents. So resistance of the disconnected resistor will be 2/0.5 = 4 ohm.
- Parallel combination of three 3 ohm resistors, connected in series with parallel combination of two 2 ohm resistors, what will be the equivalent resistance of overall combination ?
Three 3 ohm resistor are connected in parallel equivalent resistance will be 3/3=1ohm

Two 2 ohm resistor are connected in parallel equivalent resistance will be 2/2=1ohm

So, total resistance when these two combinations are series connected, the total resistance will be 1 + 1 = 2 ohm. - When a numbers of different valued resistance are connected in series, the voltage drop across each of the resistor is
Let V is the source voltage and R
_{1}, R_{2}, R_{3},........R_{n}resistances are connected in series, across the source of voltage V. Therefore, the current through the resistances will be ........... Therefore, V_{1}= IR_{1}, V_{2}= IR_{2}, V_{3}= IR_{3}.........V_{n}= IR_{n}. That means V_{n}∝ R_{n}

So, voltage drop across each resistance will be proportional to their resistive values. - All the resistances in figure shown below are 1 Ω each. The value of current in Ampere through the battery is

If look at the circuit from right side, we will see that the equivalent resistance of the circuit is 15 / 4 Ω. Current through the battery will be 1 V / (15 / 4) Ω = 4 /15 A. - Two wires A and B of same material and length l and 2l have radius r and 2r respectively. The ratio of their specific resistance will be
The resistivity of any substance depends upon its material not upon its dimensions.
- If the length of a wire of resistance R is uniformly stretched n times its original value, its new resistance is
Let's cross-section of the wire is A = πr
^{2}, length of the wire is l therefore volume of the wire is A.l. Now if the length of the wire is stretched to n times of its original length i.e. now length of the wire becomes l' = n.l. Now if r' is the new radius of the cross-section of the wire then new cross-sectional area A' = πr'^{2}. If the volume of the wire is same before and after stretching, A.l = A'.l' ⇒ πr'^{2}.n.l = πr^{2}.l ⇒ r'^{2}= r^{2}/n ⇒ πr'^{2}= πr^{2}/n ⇒ A' = A/n. Thus resistance of the wire after stretched is ρ(l'/A')= ρ{n.l/(A/n)} = n^{2}ρ(l/A) = n^{2}.R. - The resistance between the opposite faces of 1 m cube is found to be 1 Ω. If its length is increased to 2 m, with its volume remaining the same, then its resistance between the opposite faces along its length is
Volume = lengthXarea so if length is increased by 2 times then area will be decreased by 1/2, if resistance R = ρ(l/A) = 1 Ω ⇒ R' = ρ(2l/0.5A) = 4ρ(l/A) = 4R = 4 Ω.
- A wire of length l and of circular cross - section of radius r has a resistance of R ohms. Another wire of same material and of x-section radius 2r will have the same R if the length is
The cross-section of the first wire is πr
^{2}. The cross-section of the second wire is π(2r)^{2}= 4πr^{2}. The resistance of any wire depends upon the ratio of its length to area. So if the cross-section of the second wire is 4 times of that of first wire, the length of the second wire must also be 4 times of that of first if the resistance of both wires are same. - The insulation resistance of a cable of 10 km is 1 MΩ. For a length of 100 km of the same cable, the insulation resistance will be
Conductor resistance is directly proportional to length. But insulation resistance is the resistance to the flow of leakage current to ground. Since the flow of leakage current is directly proportional to the length of the conductor as because with length conductor inner and outer surface are of the insulation layer of the conductor increases. So insulation resistance is inversely proportional to the length of conductor.
- The hot resistance of the filament of a bulb is higher than the cold resistance because the temperature coefficient of the filament is >
Positive temperature coefficient refers to materials that experience an increase in electrical resistance when their temperature is raised.
- The temperature coefficient of resistance of an insulator is
.
- Four resistances 80 Ω, 50 Ω, 25 Ω and R are connected in parallel. Current through 25 Ω resistance is 4 A. Total current of the supply is 10 A. The value of R will be
The current through 25 Ω resistor is 4 A hence voltage across it is 4X25 = 100 V and this is the voltage across the supply as well as other resistors. hence current through 50 Ω and 80 Ω resistors will be 100/50 = 2A and 100/8 = 1.25 A. Therefore current through R
_{x}will be 10 - 4 - 2 - 1.25 = 2.75 A and then R_{x}= 100/2.75 = 36.36 Ω - Three parallel resistive branches are connected across a DC supply. What will be the ratio of the branch current I
_{1}:I_{1}:I_{1}if the branch resistances are in the ratio R_{1}:R_{2}:R_{3}:: 2:4:6Current is inversely proportional to resistance. Then 1/2:1/4:1/6=3:3/2:1=6:3:2. - Two resistors R
_{1}and R_{2}given combined resistance of 4.5 Ω when in series and 1 Ω when in parallel, the resistance areWhen in series

R_{1}+ R_{2}= 4.5...........(1)

when in parallel

(R_{1}*R_{2})/(R_{1}+ R_{2}) = 1

(R_{1}*R_{2})/4.5 = 1

R_{1}*R_{2}= 4.5..........(2)

COMBINING (1) AND (2),WE GET,

R_{1}= 1.5 or 3 and R_{2}= 3 or 1.5 - When a resistor R is connected to a current source, it consumes a power of 18 W. When the same R is connected to a voltage source having the same magnitude as the current source, the power absorbed by R is 4.5 W. The magnitude of the current source and the value of R are
For resistance R, connected to the current source, the consumed power is 18w i.e 18 = I
^{2}R (1) and for second condition 4.5 = V^{2}/R (2) and current and voltage having same magnitude that is V = I (3).

By solving these 3 equations we get R = 2 ohms and I = 3 A - When all the resistances in the circuit are of 1 Ω each, the equivalent resistance across the points A and B will be

All the resistances are same. There is no potential difference between central vertical resistance [like Wheatstone Bridge], so it can be imagined that it is opened. Then the equivalent resistance between A & B is :

R_{eq}= (1+1) || (1+1) || 1 Ω

R_{eq}= 2 || 2 || 1 Ω

R_{eq}= 0.5 Ω - Resistivity of metals is expressed in terms of