50 questions · timed · auto-graded
P = v2/R = I2R = VI Option (b) does not represent electrical power.
Explanation:
A fuse is a safety device that provides overcurrent protection, to an electrical circuit.
Its essential component is a metal wire or strip that melts when too much current flows through it, interrupting the current that it connects.
Therefore the working of the fuse depends upon its resistance and current flowing through it.
Explanation:
It is the SI unit of electric current.
Explanation:
In the given group, iron, tin and steel are electric conductors while glass is an electric insulator.
Hence, glass does not belong to the group formed by others.
Explanation:
The coulomb (symbol: C) is the International System of Units (SI) unit of electric charge. It is the charge (symbol: Q or q) transported by a constant current of one ampere in one second.
Explanation: Work done × Charge × Potential difference
⇒ Work done = (Current × Time) × Potential difference $[\because$ Charge = Current × Time$]$
⇒ Potential difference $=\frac{\text{Work done}}{\text{Current}\times\text{Time}}$
Explanation:
It is the temperature of an electric bulb filament when it glows.
Explanation:
An electric iron may use a power of 850 watts.
Explanation:
Wood is a good insulator of electricity. Copper, steel, and iron are metals that are good conductors of electricity.
Therefore, the wooden ruler is an electrical insulator, and all the other options in the question are electrical conductors.
Explanation:
This is so because in this case, the resistance of the wire will be low.
Explanation:
An electric current is a flow of electric charge. In electric circuits, this charge is often carried by moving electrons in a wire.
It can also be carried by ions in an electrolyte, or by both ions and electrons such as in a plasma.
It is the rate of charge flow past a given point in an electric circuit, measured in Coulombs/ second which is named Amperes.It is the charge per unit time.
Explanation:
Electricity bill is measured in units where 1 unit is equal to kWh which is the unit of energy.
Hence, when we pay for our electricity bill, we pay for the amount of electrical energy consumed by us.
Explanation:
The resistance of each bulb
$=\frac{\text{v}^2}{\text{p}}=\frac{(200)^2}{60}\Omega$
When three bulbs are connected in series their resultant resistance
$= 3\text{x} (200)^2/60$
Thus power drawn by bulb when connected across 200V supply
$\text{p}=\frac{\text{v}^2}{\text{R}_\text{re}}=\frac{(200)^2}{3\times(200)^2/60}=20\text{w.}$
Explanation:
$\text{R}=\rho\frac{\text{I}}{\text{A}}$
When the diameter is doubled, d = 2d
Radius, r' = 2r
Area of cross section, $\text{A}'=\pi\text{r}'^2=\pi(2\text{r})=4\pi\text{r}^2=4\text{A}$
The area of cross-section will increase by four times.
Then the new resistance, $\text{R}'=\frac{\rho\text{I}}{\text{A}'}$
$\text{R}'=\frac{\rho\text{I}}{4\text{A}}$
$\text{R}'=\frac{\text{R}}{4}$
Thus, the resistance will get reduced by four times.
Explanation:
We know that the resistance of a conductor is given by:
$\text{R}=\rho\frac{\text{I}}{\text{A}}$
where $\rho=\text{resistivity}$
l = length of the conductor
A = area of the cross-section of the conductor
Let the new resistance be R' when the area of cross-section of the conductor is doubled
$\text{R}'=\rho\frac{\text{I}}{2\text{A}}$
Explanation:
Power, P = VI = 12V × 0.5A = 6W
Explanation:
The number of coulombs passing through the resistor is the current passing through it.
$\text{Current}=\frac{\text{Voltage}}{\text{Resistance}}$
$\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=\frac{12}{4}$
$\text{I}={3\text{A}}$
Thus, when a $4 Ω$ resistor is connected across the terminals of a 12V battery, the number of coulombs passing through the resistor per second will be 3.
Explanation:
According to Joule's heating effect of current, heat energy is generated across a conductor due to the current flowing through it and hence its temperature increases.
Explanation:
V = 20 volts
R = 1 Ω
Power dissipated in it $\text{P}=\frac{\text{V}^2}{\text{R}}=\frac{20^2}{1}=400\text{w}$
Explanation:
The total length of the wires will affect the amount of resistance.
The longer the wire, the more resistance that there will be.
There is a direct relationship between the amount of resistance encountered by charge and the length of wire it must traverse.
After all, if resistance occurs as the result of collisions between charge carriers and the atoms of the wire, then there is likely to be more collisions in a longer wire.
More collisions mean more resistance.
Hence, when the length of the wire is doubled, the resistance of the wire is increased twice.
Explanation:
Put them all in parallel-
$\frac{1}{\text{R}}$ = $1\frac{1}{5}$ + $1\frac{1}{5}$ + $1\frac{1}{5}$ + $1\frac{1}{5}$ + $1\frac{1}{5}$
$= 5 + 5 + 5 + 5 + 5 = 25 $
$\text{R} = \frac{1}{25}\text{Omhs}$
Explanation: When resistors are connected in series then we get the maximum resistance out of the combination. When the given resistors are connected in series, the resistance of combination can be calculated as follows:
$\text{R}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{5}{5}=1\Omega$
| V1 | V2 | V3 | |
| (a) | 1.0 | 2.0 | 3.0 |
| (b) | 5.0 | 10.0 | 15 |
| (c) | 5.0 | 2.5 | 1.6 |
| (d) | 4.0 | 3.0 | 2.0 |
Explanation:
Because V = IR, the net voltage can be obtained by multiplying current with resistance.
Explanation:
This is so because if the switch, S2 is open, the current will flow through lamp A because it is a parallel circuit.
Explanation: Electric power = voltage × current
SI Unit of voltage = Volt
SI Unit of current = Ampere
So, unit of electric power is also given by, volt ampere.
Explanation:
Ratio of cross-sectional areas of the wires = 3 : 1 and resistance of thick wire $(\text{R}_{1})10\Omega$
Resistance $(\text{R})=\rho\frac{1}{\text{A}}\propto$
Therefore $\frac{\text{R}_{1}}{\text{R}_{2}}=\frac{\text{A}_{2}}{\text{A}_{1}}=\frac{1}{3}$
or $\text{R}_{2}=3\text{R}_{1}=3\times10=30\Omega$
and equivalent resistance of these two resistances in series combination
$=\text{R}_{1}=\text{R}_{2}=30+10=30\Omega$
Explanation: In series combination of resistor, the current through both the resistor are same but potential difference across each will be different.
In parallel combination current across each resistor will be different but the potential difference will be same.
The equivalent resistance can be calculated as.
$\text{R}_{\text{eq}}=\frac{4.2}{4+2}+0.5=\frac{11}{6}\Omega$
$\text{R}_{\text{eq}}=\frac{11}{6}\Omega$
The net current in the circuit is given as $\text{I}=\frac{\text{V}}{\text{R}}$
$=\frac{3}{\big(\frac{11}{6}\big)}$
$=\frac{18}{11}\text{A}, \text{so},$
$\text{I}_{\text{net}}=\frac{18}{11\text{A}}$
The circuit can be drawn as shown aside,
The voltage drop across the parallel resistance combination is $\frac{24}{11\text{V}}$ so net current in the $4\Omega$ resistor is $\frac{6}{11\text{A}}=0.55\text{A}$
Explanation:
In a series connection, the current through each device remains the same. Therefore, the current through the 40W bulb will also be 1A.
Explanation:
$2+2=4\Omega$
this combination is parallel to the third resistance. Hence effective resistance is
$4\times\frac{2}{4}+2=\frac{8}{6}=\frac{4}{3}$