Determine the following quantities for the circuits shown below:
(a) the equivalent resistance
(b) the total current from the power supply
(c) the current through each resistor
(d) the voltage drop across each resistor and
(e) the power dissipated in each resistor​.
WILL MARK AS BRAINLIEST!​

in a parallel circuit equivalent resistor is,

1/R = 1/20 + 1/100 + 1/50

so 1/R = 5/100 + 1/100 + 2/100

= 8 /100

1/R = 8/100

8R = 100

R = 12.5 ohms

b) v = IR

125 = I × 12.5

I = 125/12.5

I = 10A

c) resistor with 20 ohms

v = ir

125 = 20I

I = 6.25 A

the resistor with 100

125 = 100I

I = 125/100

= 1.25A

the resistor with 50 ohms

125 = 50I

I = 125 /50

2.5A

if u want to make sure all these calculations are crct add them up and see

A resistor is an electrical component that provides electrical resistance or limits the flow of current in a circuit. For calculating resistance (R) values of voltage(V) and current (I) should be known.

The correct quantities are (a) 12.5 ohms, (b) 10A (c)  6.25 A, 1.25A, 2.5A (d) 125 V  (e) 781.25 W, 156.25 W and 312.5 W.

The calculations are as follows:

Given,

(a) Equivalent resistance in parallel resistor:

[tex]\rm \dfrac{1}{Req} = \dfrac{1} {R1} + \dfrac{1} {R2} + \dfrac{1} {R3}[/tex]

[tex]\rm \dfrac{1}{Req} = \dfrac{1} {20} + \dfrac{1} {100} + \dfrac{1} {50}[/tex]

So,

[tex]\rm \dfrac{1}{Req} = \dfrac{5} {100} + \dfrac{1} {100} + \dfrac{2} {100}[/tex]

[tex]\rm \dfrac{1}{Req} = \dfrac{8} {100}[/tex]

Req = 12.5 ohms

(b) Current can be calculated using the formula:

V = IR

Where,

V = 125 V

I = ?

R = 12.5 ohms

Substituting values in equation:

[tex]\rm 125 = I \times 12.5[/tex]

[tex]\rm I = \dfrac {125 }{12.5} \\\\I = 10 A[/tex]

Therefore, current is 10 Amperes.

(c) Current through each resistor:

Given,

V = 125 V

I = ?

[tex]\rm 125 = I \times 20\\\\I1 = 6.25 A[/tex]

[tex]\rm 125 = I \times 100\\\\I2= 1.25A[/tex]

[tex]\rm 125 = I \times 50 \\\\I3= 2.5 A[/tex]

(d) Voltage drop across each resistor in a parallel circuit can be calculated by:

For parallel:

Given from (c)

I1 = 6.25

I2= 1.25 A

I3 = 2.5 A

[tex]\rm V1 = 6.25 \times 20\\V 1 = 125[/tex]

[tex]\rm V2 = 1.25 \times 100\\ V2= 125[/tex]

[tex]\rm V3 = 2.5 \times 50\\ V3= 125[/tex]

(e) Power dissipated by each source can be calculated by the following formula:

Given,

V = 125 V

[tex]\rm P = \dfrac {V^{2}}{ R}[/tex]

Resistor 1:

[tex]\rm P = \dfrac {(125)^ 2 }{ 20}\\\\P1 = 781.25 W[/tex]

Resistor 2:

[tex]\rm P = \dfrac {(125)^ 2 }{ 100}\\\\P2 = 156.25 W[/tex]

Resistor 3:

[tex]\rm P = \dfrac {(125)^ 2 }{ 50}\\\\P3 = 312.5 W[/tex]

Therefore, (a) 12.5 ohms, (b) 10A (c)  6.25 A, 1.25A, 2.5A (d) 125 V (e) 781.25 W, 156.25 W and 312.5 W.

To learn more about resistors in parallel circuit follow the link:

https://brainly.com/question/25686930