Question

Show how would you join three resistors, each of resistance 9 Ω so that the equivalent resistance of the combination is (i) 13.5 Ω, (ii) 6 Ω?OR(a) Write Joule's law of heating.(b) Two lamps, one rated 100 W; 220 V, and the other 60 W; 220 V, are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220 V.

Answer 1

1. equivalent resistance = 13.5 Ω
Join two resistors in parallel and one in series as shown in 1st figure
We know, in parallel ,
1/R' = 1/R₁ + 1/R₂
Here , R₁ = R₂ = R₃ = 9Ω
Now, 1/R' = 1/9Ω + 1/9Ω ⇒ R' = 9/2Ω = 4.5 Ω
So, equivalent resistance , Req = R' + R₃ [ ∵ R' and R₃ in series ]
∴ Req = 4.5Ω + 9Ω = 13.5Ω

2. Equivalent resistance = 6Ω
two resistors are in series which is parallel with 3rd resistor as shown in 2nd figure .
now, R' = R₁ + R₂ = 9Ω + 9Ω = 18Ω [ R₁ and R₂ are in series ]
now, equivalent resistance, Req = R'R₃/(R' + R₃) [ R' and R₃ are in parallel ]
∴ Req = 18 × 9/(18 + 9) = 18 × 9/27 = 6Ω

(a) Joule's Law Of Heating

When current I flow through a resistor of electric resistance R, heat is produced. The heat produced H depends directly on the square of the current, resistance and time t for which the current is allowed to pass. H=I2Rt, Is called Joule's law of heating.

(b) Power = P = 100 W
Voltage = V = 220v
Resistance = R

P = V²/R
100 = 220 ×220/R
R  = 220 × 220/100
     = 484 Ω
==========================
Power = P = 60 W
Voltage = V = 220v
Resistance = R

P = V²/R
60 = 220 × 220/R
R = 220× 220/60
    = 806.7 Ω
=========================
As the resistors are connected in parallel , 
total resistance = 1/R

1/R = 1/484 + 1/806.7
      =  806.7 + 484/484×806.7
      = 1290.7/390442.8
R = 390442.8/ 1290.7 = 302.5 ohms
    
  Total resistance = 302.5 Ω


I = current

V = IR
220 = I × 302.5

I = 220/302.5
   = 0.73 A

The current drawn is 0.73 A

Answer 2

here's UR Answer ✍️✍️✍️


1. equivalent resistance = 13.5 Ω
Join two resistors in parallel and one in series as shown in 1st figure
We know, in parallel ,
1/R' = 1/R₁ + 1/R₂
Here , R₁ = R₂ = R₃ = 9Ω
Now, 1/R' = 1/9Ω + 1/9Ω ⇒ R' = 9/2Ω = 4.5 Ω
So, equivalent resistance , Req = R' + R₃ [ ∵ R' and R₃ in series ]
∴ Req = 4.5Ω + 9Ω = 13.5Ω

2. Equivalent resistance = 6Ω
two resistors are in series which is parallel with 3rd resistor as shown in 2nd figure .
now, R' = R₁ + R₂ = 9Ω + 9Ω = 18Ω [ R₁ and R₂ are in series ]
now, equivalent resistance, Req = R'R₃/(R' + R₃) [ R' and R₃ are in parallel ]
∴ Req = 18 × 9/(18 + 9) = 18 × 9/27 = 6Ω

(a) Joule's Law Of Heating

When current I flow through a resistor of electric resistance R, heat is produced. The heat produced H depends directly on the square of the current, resistance and time t for which the current is allowed to pass. H=I2Rt, Is called Joule's law of heating.

(b) Power = P = 100 W
Voltage = V = 220v
Resistance = R

P = V²/R
100 = 220 ×220/R
R  = 220 × 220/100
     = 484 Ω

Power = P = 60 W
Voltage = V = 220v
Resistance = R

P = V²/R
60 = 220 × 220/R
R = 220× 220/60
    = 806.7 Ω



As the resistors are connected in parallel , 
total resistance = 1/R

1/R = 1/484 + 1/806.7
      =  806.7 + 484/484×806.7
      = 1290.7/390442.8
R = 390442.8/ 1290.7 = 302.5 ohms
    
  Total resistance = 302.5 Ω


I = current

V = IR
220 = I × 302.5

I = 220/302.5
   = 0.73 A

The current drawn is 0.73 A


HOPE you understand ❤️❤️