Tag: electric current, potential difference and resistance

Questions Related to electric current, potential difference and resistance

If 1 mole of electrons contains $ 6.02 \times 10^{23} $ electrons, the value of faraday constant, F, in coul/mole is :

  1. $ \frac {96320}{2} $

  2. $ 96320 \times 2 $

  3. 96320

  4. $ 6.6 \times 10^{19} $


Correct Option: A

The number of electrons striking the screen of CR is $7.5\times 10^{15}$ in 10 s. Calculate the electric current

  1. $1.2\times10^{-4}$

  2. $7.5\times10^{-4}$

  3. $1.6\times10^{-4}$

  4. $1.2\times10^{-5}$


Correct Option: A

In a conductor, $6.25\, \times\, 10^{16}$ electrons flow from its end A to B in 2 s. Which is the current flowing through conductor? $(e\, =\, 1.6\, \times\, 10^{-19}\, C)$  

  1. $5 mA$ from B to A.

  2. $5 mA$ from A to B.

  3. $10 mA$ from A to B.

  4. $10 mA$ from B to A.


Correct Option: A
Explanation:

In the question, it is given that the charge of an electron is $e=1.6\times { 10 }^{ -19 }\quad C$.

Therefore, the total charge flowing through the wire in 2 seconds will be given as $6.25\times { 10 }^{ 16 }\times 1.6\times { 10 }^{ -19 }\quad C\quad =10\times { 10 }^{ -3 }\quad C$.

The rate of flow of charge is therefore $\dfrac { 10\times { 10 }^{ -3 } \  C }{ 2  \ seconds\quad  } =5\times { 10 }^{ -3 }\quad C/s$.

one C/s = one amp.

Hence, the current flowing through a conductor is $0.0005 A$, that is $5 mA$.

If the electric potential difference between the two points is $V$ and resistance $R$, then the current between the points will be:

  1. $VR$

  2. $V/R$

  3. $VR^2$

  4. $V/R^2$


Correct Option: B
Explanation:

Ohm's law states that voltage drop across conductor is proportional current in it.

$ V \propto I$
$ V= IR$
The proportionality constant is $R$ which known as resistance of conductor
$I= \dfrac{V}{R}$

What is the approximate resistance setting of a rheostat in which $650 mA$ of current flows with a $150 V$ source?

  1. $230 ohm$

  2. $56 ohm$

  3. $15 ohm$

  4. $13 ohm$


Correct Option: A
Explanation:

$R=\dfrac{E}{I}$


$\implies R=\dfrac{150}{650\times 10^{-3}}$

$\implies R\approx 230\Omega$

Answer_(A)

If 18400 C of charge flows through an air conditioning unit every hour, what current does it draw?

  1. 5.11  A

  2. 5.8 A

  3. 4.1 A

  4. 6 A


Correct Option: A
Explanation:

Current is defined as charge flow per unit time .


$I=\dfrac{Q}{t}$, here $t=3600 sec$

So $I=\dfrac{18400}{3600}=5.11\ A$

An electric current of $5 A$ is divided into three branches forming a parallel combination. The length of the wire in three branches in the ratio $2:3:4$ and their diameter ratio $3:4:5$. Find the current ratio in three branches. 

  1. $1.5:2:1.3$

  2. $1.4:1.66:1.94$

  3. $1:2:1$

  4. $1:1:1$


Correct Option: B
Explanation:

Resistance $R=\rho l/A, A=\pi d^2/4$ is directly proportional to the length and inversely proportional to the Area, i.e., directly proportional to the square of the diameter.


As length are in ratio $2:3:4$ and diameter are in ratio $3:4:5$.

So the resistance are in ratio $2/3^2:3/4^2:4/5^2=2/9:3/16:4/25$

By ohm's law, current is inversely proportional to the resistance, so current is 

in ratio $9/2:16/3:25/4$

Dividing the total current $5A$ in this ratio, we get current ratios as:
$1.4:1.66:1.94$

Which of the cables, one rated 5 A and the other 15 A will be of thicker wire?

  1. 15 A cable.

  2. 5 A cable

  3. both cables

  4. nothing to say.


Correct Option: A
Explanation:

The cross-sectional area of the wires will affect the amount of resistance. Wider wires have a greater cross-sectional area. Water will flow through a wider pipe at a higher rate than it will flow through a narrow pipe. This can be attributed to the lower amount of resistance that is present in the wider pipe. In the same manner, the wider the wire, the less resistance that there will be to the flow of electric charge. When all other variables are the same, charge will flow at higher rates that is, more current through wider wires with greater cross-sectional areas than through thinner wires.
Hence, the 15 A rated wire is the thicker wire as more amount of current flow through it than the 5 A rated wire.

A cell of e.m.f. $\varepsilon$ and internal resistance $r$ is used to send current to an external resistance R. The current drawn from the cell is :

  1. $\displaystyle \frac{\varepsilon}{(R\, +\, r)}$

  2. $\displaystyle \frac{\varepsilon Rr}{R+r}$

  3. $\displaystyle \frac{\varepsilon R}{R+r}$

  4. $\displaystyle \frac{\varepsilon r}{R+r}$


Correct Option: A
Explanation:
Batteries and cells have an  internal resistance (r) which is measured in ohms. When electricity flows around a circuit, the internal resistance of the cell itself resists the flow of current and so thermal (heat) energy is wasted in the cell itself.

$\varepsilon =I(R+r)$

where,

$\varepsilon $ = electromotive force in volts, V

I = current in amperes, A

R = resistance of the load in the circuit in ohms,

r = internal resistance of the cell in ohms.

The above equation can be written as $I=\dfrac { \varepsilon  }{ R+r } $.

Hence, the expression for the current drawn from the cell is $I=\dfrac { \varepsilon  }{ R+r } $.

In an electric circuit containing a battery, the charge (assumed positive) inside the battery

  1. always goes from the positive terminal to the negative terminal.

  2. may go from the positive terminal to the negative terminal.

  3. always goes from the negative terminal to the positive terminal.

  4. does not move.


Correct Option: C
Explanation:

Inside the battery or cell, the positive charge move from negative terminal (lower potential) to positive terminal (higher potential) of the battery. While in external circuit the positive charge move from positive terminal to negative terminal of the battery.Hence correct option is C