Answer:
- Part 1 = 32 Ω
- Part 2 = 0.75 A
- Part 3 = 0.75 A
Explanation:
Part 1 :
the equivalent resistance for the resistors connected in series can be calculated by:
[tex]\rm R_{eq} = R_1 + R_2 + R_3[/tex]
According to the question we have,
[tex]\rm R_1 = 7.0 \: \Omega[/tex]
[tex]\rm R_2 = 13 \: \Omega[/tex]
[tex]\rm R_3 = 12.0 \: \Omega[/tex]
substituting the values in the formula
[tex]\rm R_{eq}= 7.0 \Omega + 13 \Omega + 12.0 \Omega [/tex]
[tex] \boxed{\rm {R_{eq} = 32 \: \Omega}}[/tex]
Hence, the equivalent resistance is 32 Ω.
Part 2:
Current in the circuit can be calculated by using the ohm's law,
According to ohm's law, V = IR
where,V is the voltage, I is the current and R is the resistance
Solving for I,
[tex] \rm I = \dfrac{V}{R}[/tex]
[tex] \rm I = \dfrac{24.0}{32}[/tex]
[tex] \boxed {\rm{I = 0.75 \: A}}[/tex]
Hence, the current flowing in the circuit is 0.75 A.
Part 3 :
We are given that A 7.0 Ω, a 13 Ω and a 12.0 Ω resistor are connected in series with a 24.0 V battery. In series, the current flowing through each resistor is same.
Hence, the current in each resistor is 0.75 A.
