6.0 resistor, a 12.0 resistor, and a 18.0 resistor are connected in series with a 24 V battery. (a) Determine the equivalent resistance for the circuit. (b) Determine the current in the circuit. A (c) Determine the current in each resistor. (d) Determine the potential difference across each resistor.

(a) R=36

(b) I=0.6

(c) I1=4 I2=2 I3=1.3

(d) v1=24 v2=24 v3=23.4

Explanation:

(a) Determine the equivalent resistance for the circuit.

R=r1+r2+r3

R=6+12+18

R=36

Now,

(b) Determine the current in the circuit.

V=IR

I=V/R

I=24/36

I=0.6

(c) Determine the current in each resistor.

For first resistor,

I1=V/r1

I1=24/6

I1=4

For second resistor,

I2=V/r2

I2=24/12

I2=2

For third resistor,

I3=V/r3

I3=24/18

I3=1.3

Now,

(d) Determine the potential difference across each resistor.

For first resistor,

v1=I1*r1

v1=4*6

v1=24

For second resistor,

v2=I2*r2

v2=2*12

v2=24

For third resistor,

v3=I3*r3

v3=1.3*18

v3=23.4

R1=6ohms

R2=12ohms

R3=18ohms

Since it is connected in series

Therefore, the equivalent resistance

Rt = R1+R2+R3

Rt=6+12+18

Rt=36ohms

(B) Determine the current

From V=IR

I=V/R

I=24/36

I=0.67A

(C) current in each resistor

I1=V/R1

I1=24/6

I1=4A

I2=V/R2

I2=24/12

I2=2A

I3=V/R3

I3=24/18

I3=1.33A