Answer to Question #264658 in Electric Circuits for lara

Question

a) A 200Ω resistor is connected in series with other resistor of 2kΩ and a battery of 150V. Determine the equivalent resistance and the current flowing through the circuit. 

b) A 12V battery is connected in a series circuit with a resistor R1 = 8Ω and a resistor R2 = 3Ω, determine the current that passes through R1 and R2. 

Answer 1

Solution;

(a)

Equivalent resistance is;

$R_e=R_1+R_2=200+2000=2200\Omega$

Current flowing through the circuit;

$I=\frac{V}{R_e}=\frac{150}{2200}=0.06818A$

(b)

Equivalent resistance;

$R_e=R_1+R_2=3+8=11\Omega$

We know;

$I=I_1=I_2$

$I=\frac{V}{R_e}=\frac{12}{11}=1.09A$

Hence;

$I_1=1.09A$

$I_2=1.09A$