Question

Three resistors, of 40 Ω, 60 Ω, and 120 Ω, are connected in parallel, and this parallel group is connected in series with 15 Ω in series with 25 Ω. The whole system is then connected to a 120 V source. Determine (a) the current in the 25 Ω, (b) the potential drop across the parallel group, (c) the potential drop across the 25 Ω, (d) the current in the 60 Ω, (e) the current in the 40 Ω.

Answer 1

(a) The current in the 25 Ω resistor
`(\( I_(25) \)) is \( 2.4 \, \text{A} \).`

(b) The potential drop across the parallel resistors
`(\( V_{\text{parallel}} \)) is \( 24 \, \text{V} \).`

(c) The potential drop across the 25 Ω resistor
`(\( V_(25) \)) is \( 60 \, \text{V} \).`

(d) The current in the 60 Ω resistor
`(\( I_(60) \)) is \( 0.4 \, \text{A} \).`

(e) The current in the 40 Ω resistor
`(\( I_(40) \)) is \( 0.6 \, \text{A} \).`

How did we get the values?

Find the equivalent resistance for the parallel resistors
`(\( R_{\text{parallel}} \))`:

`\[ \frac{1}{R_{\text{parallel}}} = (1)/(40) + (1)/(60) + (1)/(120) \]`

Calculating this gives:

`\[ \frac{1}{R_{\text{parallel}}} = (1)/(40) + (1)/(60) + (1)/(120) \]`

`\[ \frac{1}{R_{\text{parallel}}} = (6 + 4 + 2)/(120) \]`

`\[ \frac{1}{R_{\text{parallel}}} = (12)/(120) \]`

`\[ R_{\text{parallel}} = (120)/(12) = 10 \, \Omega \]`

Find the total resistance
`(\( R_{\text{total}} \))` for the circuit:

`\[ R_{\text{total}} = R_{\text{parallel}} + R_4 + R_5 = 10 + 15 + 25 = 50 \, \Omega \]`

Find the current in the circuit
`(\( I_{\text{total}} \))` using Ohm's Law:

`\[ I_{\text{total}} = \frac{V}{R_{\text{total}}} = (120)/(50) = 2.4 \, \text{A} \]`

Find the potential drop across the parallel resistors
`(\( V_{\text{parallel}} \)) using Ohm's Law:`

`\[ V_{\text{parallel}} = I_{\text{total}} \cdot R_{\text{parallel}} = 2.4 \cdot 10 = 24 \, \text{V} \]`

Find the potential drop across the 25 Ω resistor
`(\( V_(25) \))` using Ohm's Law:

`\[ V_(25) = I_{\text{total}} \cdot R_5 = 2.4 \cdot 25 = 60 \, \text{V} \]`

Find the current in the 25 Ω resistor
`(\( I_(25) \))` using Ohm's Law:

`\[ I_(25) = (V_(25))/(R_5) = (60)/(25) = 2.4 \, \text{A} \]`

Find the current in the 60 Ω resistor
`(\( I_(60) \))` using Ohm's Law:

`\[ I_(60) = \frac{V_{\text{parallel}}}{R_2} = (24)/(60) = 0.4 \, \text{A} \]`

Find the current in the 40 Ω resistor
`(\( I_(40) \))` using Ohm's Law:

`\[ I_(40) = \frac{V_{\text{parallel}}}{R_1} \\= (24)/(40) \\= 0.6 \, \text{A} \]`

Consequently,

(a) The current in the 25 Ω resistor
`(\( I_(25) \)) is \( 2.4 \, \text{A} \).`

(b) The potential drop across the parallel resistors
`(\( V_{\text{parallel}} \)) is \( 24 \, \text{V} \).`

(c) The potential drop across the 25 Ω resistor
`(\( V_(25) \)) is \( 60 \, \text{V} \).`

(d) The current in the 60 Ω resistor
`(\( I_(60) \)) is \( 0.4 \, \text{A} \).`

(e) The current in the 40 Ω resistor
`(\( I_(40) \)) is \( 0.6 \, \text{A} \).`