Answer to Question #196979 in Mechanics | Relativity for jireh mae

Question

SERIES CIRCUIT:

1. Two 80 Ohms are connected in series with 0.15A flowing through them?
FIND;
A. total resistance
B. voltage drop across resistors
C. current through the circuit

2. in this circuit, the voltage drop across the upper resistor is 4.5V.
What is the battery voltage?
What is the current through the circuit?
What is the resistance in the circuit?
(30 ohms in the upper side of the circuit and 50 ohms in the right side of the circuit)

Answer 1

Gives

$R_1=R_2=80 \Omega$

Current $(i)=0.15A$

Total resistance

Series combination

$R_{net}=R_1+R_2$

$R_{net}=80+80=160\Omega$

Voltage drop across the resistance

$V_1=iR_1$

$V_2=iR_2$

$V_1=V_2=0.15\times80=12V$

Net current through the circuit

$I=\frac{V}{R_{net}}$

$I=\frac{24}{160}=0.15A$

Part (2)

Resistance of parrelled combination

$R_1=30\Omega$

$R_2=50\Omega$

Voltage drop across R₁ resistance

V₁=4.5V

Net resistance

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$R=\frac{R_1R_2}{R_1+R_2}$

$R=\frac{30\times50}{30+50}=18.75\Omega$

R_net=18.75 ohm

Current uper resistance

$V=i_1R_1$

$i_1=\frac{V}{R_1}$

$i_1=\frac{4.5}{30}=0.15A$

Current flow R₂ resistance

$i_2=\frac{V}{R_2}$

$i_2=\frac{4.5}{50}=0.09A$

Battery voltage (v)= 4.5V

Parrelled combination voltage drop same each resistance

Current through the circuit

$I=i_1+i_2$

$I=0.15+0.09=0.24A$

$I=0.24A$