Science M.C.Q

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.

The resistors of $1 Ω, 2 Ω$ and $3 Ω$ are connected in series. Therefore, the net resistance,
$\text{R} = \text{R}_1 + \text{R}_2 +\text{R}_3$
$\text{R}=1\Omega+2\Omega+3\Omega=6\Omega$
Current in the circuit will be,
$\text{I}=\frac{\text{V}}{\text{R}}$
or $\text{I}=\frac{2}{6}=\frac{1}{3}\text{A}$
Current $=\frac{1}{3}\text{A}$
Therefore, the voltage across the 3I resistor,
V = IR
or $\text{V}=\frac{1}{3}\times3=1\text{V}$

$\text{R}=\rho\frac{\text{I}}{\text{A}}$
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.

Voltage, V = 12V
Current, I = 0.5A
Resistance, $\text{R}=\frac{\text{Voltage}}{\text{Current}}$
$\text{R}=\frac{12}{0.5}$
$\text{R} = 24Ω$
Thus, if a car's headlight bulb working on a 12V car battery draws a current of 0.5A, the resistance of the light bulb will be $ 24Ω$

If the resistance wire is cut into five pieces, the resistance of each wire is $\frac{\text{R}}{5}.$ If we connected the pieces in parallel, we will get the net resistance as $\frac{\text{R}}{25}$ Therefore, the ratio will be $25.$

When $25 Ω$ and $15 Ω$ are connected in series, then:
Total resistance, R = 25 + 15 = 40I
This 40I is connected in parallel with the 40I resistor.
Therefore, the net resistance $=\frac{40}{2}=20\text{I}$

El Total power provided in the house, P = VI = 100A × 220V = 22000W
This is the maximum power that can be drawn in the house.
Total power drawn by x bulbs = 75 x W
Maximum power x bulbs can draw = 22000W.
$\Rightarrow75\text{x}=22000$
$\Rightarrow \text{x}=\frac{22000}{75}\approx293$

Because $V = IR$, the net voltage can be obtained by multiplying current with resistance.

Resistance = $25Ω$
Voltage = 230V
$\text{Current}=\frac{\text{Voltage}}{\text{Resistance}}$
$\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=\frac{230}{25}$
$\text{I}=9.2\text{A}$
Thus, if an electrical appliance has a resistance of $25Ω$ and when that electrical appliance is connected to a 230V supply line, the current passing through it will be 9.2A.

Since $\text{P}=\frac{\text{V}^2}{\text{R}}, $ when the potential difference is halved, the power becomes one-fourth.

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
$\text{R}'=\rho\frac{\text{I}}{2\text{A}}$

The resistivity of a certain material is $0.6 Ω\ \text{m}. $The material is most likely to be a semiconductor because it has moderate resistivity.

Here, current, I = 5A
Time, t = 10s
So, using the relation, $\text{I} =\frac{\text{Q}}{\text{t}}$
Q = I × t
Charge, Q = 5 × 10 = 50C
Thus, a charge of 50C is passed in 10s.

Current drawn by the kettle, $\text{I}=\frac{\text{P}}{\text{V}}=\frac{3000}{230}=13\text{A}$
So the cable should be able to carry more current than this and hence the answer is 15A.

Resistance of the wire is given by:
$\text{R}=\rho\frac{\text{I}}{\text{A}}$
When the diameter is halved:
$\text{d}'=\frac{\text{d }}{2}$
Radius:
$\text{r}'=\frac{\text{r}}{2}$
Area of cross-section:
$\text{A}'=\pi\text{r}'^2=\pi\Big(\frac{\text{r}}{2}\Big)^2=\frac{\pi\text{r}}{4}=\frac{\text{A}}{4}$
Area of cross-section will get reduced by four times.
Then the new resistance:
$\text{R}'=\frac{\rho\text{I}}{\text{A}}$
$\text{R}'=\frac{4\rho\text{I}}{\text{A}}$
$\text{R}'=4\text{R}$
Thus, the resistance will increase by four times.