WBBSE Class 10 Physical Science Chapter 6 Current Electricity Long Answer Questions

Class 10 Physical Science Solution WBBSE Chapter 6 Current Electricity Broad Answer Type Questions

Question 1. State Joule’s laws of heating effect of current.
Answer:

Joule’s Laws (1841):

1.  First Law:

The amount of heat produced in a conductor in a given interval of time of proportion to the square of the current passed. Thus if H be the amount of heat generated in a conductor having resistance R when current 1 passes through it in time t, then

⇒ Hα I² (When R and t are kept constant)

2. Second law:

The amount of heat produced by a given current in a given time is proportional of the resistance of the conductor.

⇒  H α R (When I and t are kept constant)

3. Third Law:

The amount of heat produced in a given conductor by a given current is proportional to the time for which the current passes.

⇒  Hα t (When I and R are kept constant)

Combining the three laws, we have:

⇒ H α I² RT (When 1, R and t vary) or,

H = \(\frac{I^2 R T}{J}\)

IRT (J= mechanical equivalent of heat = 4.2 joule (calorie)

If I am in ampere, R in ohm, t in second, and H in calorie, then

H= \(\frac{I^2 R T}{J}\)

H= 0.241 RT calorie.

Question 2. Short Note: 

1. Ohmic resistance
2. Non-ohmic resistance

Answer:

1. Ohmic Resistance :

The resistors, which obey Ohm’s Law are said to have ohmic resistances.

Example: All metal or metallic alloy.

2. Non-ohmic resistance :

The resistors which do not obey Ohm’s law are said to have non-ohmic resistance.

Example: Electronic value.

Question 3. Two resistors of 30W and 60W are connected in parallel in an electric circuit. How does the current passing through the two resistors compare?
Answer:

Given

Two resistors of 30W and 60W are connected in parallel in an electric circuit

The potential difference across 30W = B potential difference across 60W

i.e I₁ R₁= I₂ R₂ Or,

⇒ \(\frac{I_1}{I_2}=\frac{R_2}{R_1}\)

= \(\frac{60 \Omega}{30 \Omega}\)

= 2

Question 4. Short Note-Direct Current.
Answer:

Direct Current:

If a resistor connects with the two terminals of an electric cell, then through the resistor a steady current of constant magnitude flows in the same direction (from the positive pole of the cell to the negative pole) and this is called a direct current.

Class 10 Physical Science Solution WBBSE

Question 5. A resistance of 6 Ω is connected with a cell of em. f. 1.5 and negligible internal resistance calculate the current flowing through it.
Answer:

Given

A resistance of 6 Ω is connected with a cell of em. f. 1.5 and negligible internal resistance

V= IR Or, I= \(\frac{V}{R}\)

= \(\frac{1.5}{6}\)

= \(\frac{1}{4}\)

V = 0.25 Amp

The current flowing through it is 0.25 Amp

Question 6. Find the specific resistance of the material of a wire of length 100cm, area of cross-section 0.2 cm,^2, and resistance 2 ohms.

Answer:

R = P× \(\frac{1}{\mathrm{~A}}\)

∴ P= \(\frac{\mathrm{RA}}{\mathrm{1}}\)

= \(\frac{20 \times 0.2}{100}\)

= 40 ×10^-4  Ohm^-cm

So,

R = 20hm

1 = 100 cm

A = 0.2 cm²

P= ?

Question 7. Find effective resistance of the resistors 2 ohm, 4 ohm, 5 ohm connected in

1. Series
2. Parallel.

Answer:

We know, for series combination equivalent resistance

R= r₁+ r₂+r₃

= 2+4+5

= 11 ohm.

We also know, for parallel combination equivalent resistance.

⇒ \(\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\)

= \(\frac{1}{2}+\frac{1}{4}+\frac{1}{5}\)

= \(\frac{19}{20}\)

∴ \(\frac{1}{R}\)= \(\frac{19}{20}\) Or,

= \(\frac{20}{19}\)

R = 1.05 Ohm

Question 8. Short Note-Electric Circuit.
Answer:

Electric Circuit: A continuous conducting path between the terminals of a source of electricity, is called an electric circuit.

Open electric → An electric circuit in which the flow of current stops, because of an open switch is called an open electric circuit. Closed electric

Open Electric Circuit:

Closed Electric Circuit:

Circuit → An electric circuit in which a current flows continuously, because the switch is closed is called a closed electric circuit.

Question 9. There are two copper wires of equal length. The radius of one is twice the other. Find the ratio of their resistances.

Answer:

Given

There are two copper wires of equal length. The radius of one is twice the other.

We know, R= p \(\frac{1}{A}\)

= \(p \frac{1}{\lambda r^2}\)

(r= radius of the wire)

Since, the length and material are the same.

⇒ \(\mathrm{R} \alpha \frac{1}{\mathrm{r}^2}\)

i.e. \(\frac{R_1}{R_2}=\frac{r_2^2}{r_1^2}=\frac{\left(2 r_1\right)^2}{r_1^2}\)

= 4

∴ The thinner wire has a resistance four times the resistance of the thicker wire.

Question 10. Establish the relation between the emf, terminal voltage and internal resistance.
Answer:

The relation between the emf, terminal voltage and internal resistance

Let, a cell of emf E and internal resistance r is connected to an external resistance R

The total resistance of the circuit = R+r

So, the current drawn from the cells

I= \(\frac{e . m . f \text { of the cell }}{\text { total resistance }}\)

I= \(\frac{E}{R+r}\)

E= I (R+r)

The terminal voltage of the cell, V = IR voltage (v) drops due to internal resistance -I As the work is done carrying a unit positive charge once through a complete circuit,

E = V + v

v = E-V

Internal resistance r = \(\frac{V}{I}\)

=  \(\frac{E-V}{I}\)

= \(\frac{E-V}{V / R}\)

= \(R \cdot\left[\frac{E}{V}-1\right]\)

Question 11. A current of 0.5 ampere passes through a wire of resistance 2.5 ohm. for 1 hour. Find the heat produced.
Answer:

Given

A current of 0.5 ampere passes through a wire of resistance 2.5 ohm. for 1 hour.

We know,

H  = \(\frac{I^2 R T}{J}\)

= \(\frac{I^2 R T}{4^2}\)

∴ H \(\frac{(0.5)^2 \times 2.5 \times 3600}{4^2}\)

= 535.7 caloric

So,

I = 0.5 ampere

R = 2.5 ohm

t = 1hr.

= 3600 sec.

H =?

Question 12. The resistance of a wire of cross-section area 0.01 cm2 is 10 ohm. What is the length of the wire? The specific resistance of the wire is 50 × 10-6 ohm-cm.
Answer:

Given

The resistance of a wire of cross-section area 0.01 cm2 is 10 ohm.

We know,

R = p = \(\frac{\ell}{\mathrm{A}}\) Or,

l= \(\frac{\mathrm{RA}}{p}\) Or,

l= \(\frac{10 \times 0.01}{50 \times 10^{-6}}\)

∴ l= 2000 cm

So,

R= 10 Ohm

A= 0.01 cm²

p= 50 × 10^-6

Ohm^–cm

1= ?

Question 13. What is the difference between e.m.f and p.d?
Answer:

Difference between e.m.f. and p.d. : e.m.f (electromotive force):

Question 14. A resistance of 6Ω is connected with a cell of e.m.f., 1.5 V, and negligible internal resistance, Calculate the current flowing through it.

Answer:

Given

A resistance of 6Ω is connected with a cell of e.m.f., 1.5 V, and negligible internal resistance

V = IR

I = \(\frac{V}{R}\)

I = \(\frac{1.5}{6}\)

I = \(\frac{1}{4}\)

= 0.25 Amp.

Question 15. A cell of e.m.f. 1.8v is connected to an external resistance of 22 when p.d. recorded at its terminal is 1.6v. Find the internal resistance of cell.

Answer:

Given

A cell of e.m.f. 1.8v is connected to an external resistance of 22 when p.d. recorded at its terminal is 1.6v.

E = 1.8V, V 1.6 volt, R = 2.2

Internal resistance = \(\frac{R(E-V)}{V}\)

= \( \frac{2(1.8-1.6)}{1.6}\)

= 0.25.Ω