Question

Consider a circuit with two resistors in parallel R_1 = 10 ohm and R_2 = 5 ohm.A) Determine the total resistance of the circuit.B) Calculate the total current flowing through the battery.C) What are the current and voltage through each resistor?D) Repeat the calculations for a series configuration.

Answer 1

Step-by-step explanation:

Given

`R_1=10 \Omega`

`R_2=5 \Omega`

when resistance in Parallel

`(1)/(R_(p))=(1)/(R_1)+(1)/(R_2)`

`R_p=(R_1R_2)/(R_1+R_2)`

`R_p=(10)/(3)`

Suppose V is voltage of battery

Total Current
`i=(3V)/(10)`

Since Circuit is Parallel therefore Voltage across both resistor is same

`V=i_1R_1=i_2R_2`

and
`i_1+i_2=i`

`i_1+i_1\cdot (R_1)/(R_2)=i`

`i_1(1+(10)/(5))=(3V)/(10)`

`i_1=(V)/(10)`

`i_2=(2V)/(10)`

(b) When Circuit is in series

`R_s=R_1+R_2`

`R_s=10+5=15 \Omega`

since circuit is in Series therefore current is same in both resistor

Current
`i=(V)/(15) A`

Voltage drop across
`R_1=i* R_1`

`V_1=(V)/(15)* 10=(2V)/(3)`

`V_2=(V)/(15)* 5=(V)/(3)`