Question

find the effective resistance, voltage drop and current in each resistor​

Answer 1

Answer:

let

R1 = 5

R2 = 10

R3 = 2

R4 = 3

total resistance =

1/R' = 1/R1 +1/R2

R' = 3.33 ohm

1/R'' = 1/R3 +1/R4

R'' =  1.2 ohm

total R = 3.33 + 1.2

= 4 .53

Explanation:

Answer 2

Effective resistance is the total resistance of the circuit, so let's find the total resistance which is according to the diagram, is the sum of the parallel combination of resistors.

First, Let's find the total resistance of each parallel combination,

➞ 1/R₁ = 1/5 + 1/10

➞ 1/R₁ = (2+1)/10

➞ R₁ = 10/3 Ω

Now, moving to the second parallel combination, we have

⇒ 1/R₂ = 1/2 + 1/3

⇒ 1/R₂ = (3 +2)/6

⇒ R₂ = 6/5 Ω

Now, There are two resistors which are in series of resistances 10/3 Ω & 6/5Ω respectively.

So, In series, the equivalent resistance is normally the addition of resistances,

⇒ Eff. resistance = 10/3 + 6/5

⇒ Eff. resistance = (10×5 + 6×3)/15

⇒ Eff. resistance = (15 + 18) / 15

⇒ Eff. resistance = 33/15

⇒ Eff. resistance = 11/5 Ω

Moving to the second option, we have to find the voltage drop and current in each resistor.

Regarding the circuit, the total current that will flow through the each parallel combined resistors os the same. which is equal to,

⇒ V = I R

⇒ 24 = I × 11/5

⇒ I = 24×5 / 11

⇒ I = 120/11 A

Now, In first parallel combined resistor,

5 Ohm resistor:

Here, The potential difference will be equal to,

⇒ V = I R / 2 [ R = total resistance of this ]

[ divided by 2 because there are two resistors and the voltage is same accross each resistor in parallel combination ]

⇒ V = 120/11 × 10/3 / 2

⇒ V = 400/11 × 2

⇒ V = 200/11 v or 18.18 V

Because the current in parallel combined resistors is not the same, hence let's find it using the ohm's law,

5 Ω Resistor:

⇒ V = I R

⇒ 200/11 / 5 = I

⇒ 200 / 11 × 5 = I

⇒ I = 40/11 A or 3.63 A

10 Ω Resistor:

⇒ V = I R

⇒ 200/11 = I × 10

⇒ 200/11 × 10 = I

⇒ I = 20/11 A or 1.81 A

Moving to the second parallel combination of resistors where the total resistance is 6/5 Ω and similarly as of first combination, the voltage here in each resistor is the same while the current is not.

⇒ V in each resistor = (24 - 200/11) / 2

⇒ V in each resistor = 64 / 22 v or 2.90 V

2 Ω Resistor:

⇒ V = I R

⇒ 64/22 = I × 2

⇒ I = 32/22 A or 1.45 A

3 Ohm Resistor:

⇒ V = I R

⇒ 64/22 = I × 3

⇒ I = 64/66 A or 0.96 A