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Ex 6)  Series Circuit

a) Find the 'equivalent resistance'

R_T = R₁ + R₂

= 2 Ω + 1 Ω = 3 ohms

b) Find the current (I_T) going through this circuit

V_T = I_TR_T

12 Volt = I_T(3 ohms)

I_T = 4 amps

c) Find potential drop across R₁ & R₂

V = IR

V₁  = 4 amps(2 ohms)

= 8 Volts

V₂  = 4 amps(1 ohms)

= 4 Volts

Ex 7)

a) Find the combined resistance

(R_T)

V_T = I_T R_T

= 60. volts/10. amps

= 6.0 ohms

b) Find the current in R₁

V₁ = I₁R₁

60. Volts = I₁20. ohms

I₁ = 3.0 amps

c) Find I₃

V₃ = V_T = 60. volts

Use Ohm's Law

V₃ = I₃R₃

60. volts = I₃30. ohms

I₃  = 2.0 amps

d) Find R₂

e) Find the value of the second resistor

In a series circuit:

The potential difference (V) across a resistor is proportional to resistance

Ex 8)

If the voltage drop across the 3 ohms resistor is 4 volts then the voltage drop across the 6 ohm resistor is

3.0 ohms : 4.0 volts

6.0 ohms:x

x = 8.0 volts

a) Find the total voltage in this series circuit

V_T = V₁ + V₂

V_T = 8.0 V + 4.0 V

= 12 V

b) Find the combined resistance in this circuit

R_T = 6.0 ohms + 3.0 ohms

R_T = 9.0 ohms

c) Find the total current in this circuit

V_T = I_TR_T

 12 V = I_T(9.0 ohms)

I_T  = 1.3 Amps

Ex 9)

a) Compare the current through each resistor

Same (Series Circuit) 

b) What is the current in this circuit?

V_T = I_TR_T 

60. V =  I_T(30. ohms)

I_T  = 2.0 Amps

c) potential difference in the 20 ohm resistor?

V = IR

= 2 amps(20 ohms)

= 40 V

Find R₂

August Sampler

V = IR

8 V = 1 A(R_T)

R_T  = 8 ohms

R = 6 ohms

Attaching Meters