Respuesta :
Answer:
(A) R =120 ohm (b) 0.1 A (C) current through 200 ohm is 0.06 A and current through 300 ohm is 0.04 A
Explanation:
We have two resistances [tex]R_1=200ohm\ and R_2=300ohm[/tex]
(a) As the resistances are connected in parallel
So [tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
[tex]R=\frac{R_1R_2}{R_1+R_2}=\frac{200\times 300}{200+300}=120ohm[/tex]
(b) The combination is connected through 12 V battery
So current [tex]i=\frac{V}{R}=\frac{12}{120}=0.1A[/tex]
(C) According to current division rule
Current through 200 ohm resistor [tex]=0.1\times \frac{300}{200+300}=0.06A[/tex]
Now current through 300 ohm resistor = 0.1-0.06=0.04 A
Answer:
a) 120 Ω
b) 0.1 A
c) 0.06 A , 0.04 A
Explanation:
Given:
Resistance R₁ and R₂ = 200 ohms and 300 ohms
Potential difference or voltage = V = 12 V
a) When resistors are in parallel, the equivalent resistance is given by the formula [tex]\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}[/tex]
⇒ R = [tex][\frac{1}{R_{1}}+\frac{1}{R_{2}}]^{-1}[/tex]
= [tex][\frac{1}{200}+\frac{1}{300}]^{-1}[/tex] = (0.00833)⁻¹
⇒ Equivalent resistance = R = 120 ohms.
b) Current out of the battery = I = V/R = 12/120 = 0.1 A.
c) Since the resistors are in parallel, Voltage across each is equal to 12V.
Current flowing in R₁ = V/R₁ = 12 /200 = 0.06 A
Current flowing in R₂ = V/R₂ = 12 /300 = 0.04 A
