Current Electricity

A) The appliances are connected parallel to each other.

B) As the appliances are connected in parallel, the potential difference across individual appliances should be same. In India voltage ratings of all home appliances are constant i.e. 230V.

C) The current passing through each appliance won't be the same as the resistance of each appliance is different. (Current depends on both potential difference and resistance, I = V/R)

D) The domestic appliances are connected in this way as in a parallel circuit even if one of the appliance stops working, the rest would still be working but in case of series connection all the appliances stop working if one of them stops working.

E) If the TV stops working, the other appliances do not stop working as they are connected in parallel. In parallel connection, the potential difference between the loads is the same and the total current in the circuit is the sum of current flowing through each path in the circuit.

The circuit with attached symbols is given below:

Ohm’s law can be proved with the help of the above circuit.

Here we are assuming current to be constant and voltage to be varying.

A. The burning of filament is due to excessive heating when current passes through it. More the resistance of the wire, more is the heating. Therefore, the resistance of the circuit should be decreased which can be done if both the bulbs are connected in parallel.

We know that equivalent resistance in case of parallel combination of two resistances R₁ and R₂ is

In the above case,Ω which is less than the individual resistances of both the wires.

Therefore, both the bulbs should be connected in parallel.

B. Characteristics of a parallel circuit:

1) A parallel circuit has two or more paths for current to flow through and the total current is equal to sum of currents flowing through these paths.

2) Voltage is same across each component of parallel circuit.

3) The current in all the branches is same. As shown in the figure:

C. The effective resistance in the above circuit is given as:

Where R₁ and R₂ are the resistances connected in parallel.

or

A. According to the equation V = IR

R = V/I

Resistance in the 1st case R₁ = V₁/I₁

= 4/9

= 0.444 Ω

Resistance in the 2nd case R₂ = V₂/I₂

= 5/11.25

= 0.444 Ω

Resistance in the 3rd case R₃ = V₃/I₃

= 6/13.5

= 0.444 Ω

Average resistance R_avg =

= (0.444 + 0.444 + 0444)/3

= 0.444 Ω

B. In this case as R is constant, we can observe the increase in I as V increases.

We know that V = IR .

It represents a straight line starting from origin with a slope R. Therefore, the nature of graph between current and potential difference will be a straight line.

The graph is given below:

C. The graph proves Ohm's law.

According to Ohm's law the current passing through a conductor between two points is directly proportional to the voltage across the two points.

I∝V

I = V/R

We know that Resistivity ρ = RA/L

where R = Resistance of the conductor

A = Cross sectional area of the conductor

L = Length of the conductor

Here A = a and L = x

Therefore, Resistivity ρ = Ra/x

It's units are ohm.m²/m = Ωm

A. When both S₁ and S₂ are closed, R₄ is short circuited.

R₃ and (R₁||R₂) are in series

Effective resistance of the circuit R_effective =

⇒

Let the applied voltage be V.

Current in the circuit

⇒

B. When both S₁ and S₂ are open, R₃ , R₁ and R₄ are in series.

The effective resistance of the circuit,

R_effective = R₃ + R₁ + R₄

Let the applied voltage be V.

Current in the circuit I = V/R_effective

C. When S₁ is closedbut S₂ is open, R₃, (R₁||R₂) and R₄ are in series.

As S₁ is closed, R₁ and R₂ are parallel to each other.

The effective resistance of the circuit is

R_effective = R₃ + (R₁||R₂) + R₄ ( When R₁||R₂ )

=

Let the applied voltage be V

Current in the circuit I = V/R_effective

I =

A. The same current flows through all the three conductors x₁, x₂ and x₃. So, all the three must be connected in series.

Because in series connection, the current in all the wires is same.

B. Effective resistance X is larger than x₁, x₂ and x₃ individually.

The only way in which this can be possible is if all the three conductors are connected in series.

In series combination of resistances,

R_effective = R₁ + R₂ + R₃

Therefore, the effective resistance is greater than the resistances taken individually.

C. Effective resistance X is smaller than x₁ , x₂ and x₃ taken individually. This can happen only when the three resistances are connected in parallel.

In parallel combination of resistances,

Therefore, the effective resistance reduces when resistances are connected parallel to each other.

D. The potential across the conductors is same when they are connected in parallel.

E. is the formula for calculating effective resistance of a circuit when conductors are connected in series

F. is the formula for calculating effective resistance of a circuit when conductors are connected in parallel.

Given:

Resistance of a 1m(100cm) long(L₁) nichrome wire = 6 Ω (R₁)

We know that resistance is directly proportional to length of the

conductor.

Therefore R₁/R₂ = L₁/L₂

6/R₂ = 100/70

R₂ = 4.2 Ω

Given:

Effective resistance when two resistors are connected in series =80 Ω

Let us consider the resistances as R₁ and R₂ in series combination,


V = V₁ + V₂ and V=IR

As we know that in series connection current is constant, we get

R= R₁ + R₂ = 80 Ω ---(1)

Effective resistance when two resistors are connected in parallel combination = 20Ω

........(2)

putting the value of equation(1) in equation(2)

R₁R₂ = 1600

(R₁-R₂)² = (R₁ + R₂)² - 4R₁R₂

= 80² - (4×1600)

= 0

R₁-R₂ = 0 ---(3)

adding (1) and (3)

R₁₌40

Subtracting (1) and(3)

R₂=40Ω

we get R₁ = R₂ = 40Ω

Given:

Charge (q) = 420C

Time for which the charge flows (t) = 5 min = 300 s

We know that current I = q/t

= 420/300

= 1.4 A

Therefore current flowing through the conductor is 1.4 A.