Question

1. A 25 Ω resistor and a 35 Ω  resistor are connected in series and placed across a 120 V source.  (a)  Determine the effective resistance of the circuit.  (b)  What is the current in the circuit?  (c)  What is the voltage drops across the 25 Ω resistor?  (d)  What is the power used in the 35 Ω resistor?

Answer 1

Given:

A circuit with:

• 2 resistors of 35 $\Omega$ and 25 $\Omega$
• resistors connected in series combintion
• a source (battery) of 120 V

a) Effective Resistance of the circuit.

Let, R₁=35$\Omega$ and R₂ = 25$\Omega$

As we know that the resistors are in series combination,

then current flow through both the resistors will be same.

$\therefore$ equivalent resistance of the circuit will be given by

$\boxed{R_eq=R_1+R_2}$

${R_eq= 35+ 25}$

$\boxed{R_eq=60}$

b) Current in circuit

According to OHM'S LAW Current in circuit is given by:

$\boxed {I={V\over R_eq}}$

here; V=120

R_eq=60

$\boxed{\therefore I={120\over60}=2 A}$

c) Voltage drop across 25 $\Omega$ resistance

According to Ohm's law

in a series combination voltage drop across any resistor is given by

$\boxed{V_1=IR_1}$

here;

I=2A

R=25$\Omega$

$\therefore$ voltage across 25 $\Omega$ is

$\boxed{V_1=2 *25= 50 volts}$

d) power used in 35$\Omega$ resistor

In a series combination;

Power used by load or resistor is given by :

$\boxed{P_1=I^2R_1}$

here,

I=2 A

R= 35$\Omega$

$\boxed{\therefore P=(2)^2 *35= 140 W}$