Answer:
The value of JoAnna's resistors = 90 Ω
Explanation:
Given:
Circuit 1 :
Jason's circuit:
Where Jason/s circuit have three resistor.
One of the resister is, R1 = 24-Ω
Two other two resistor R2 and R3 = 12-Ω and 12-Ω
Lets find the equivalent resistance of Jason's circuit.
⇒ Equivalent resistance, [tex]R_E[/tex] = [tex]R_1 + R_e[/tex] ...equation (i)
⇒ [tex]\frac{1}{R_e} = (\frac{1}{R_2} + \frac{1}{R_3})[/tex] ...equation (ii)
⇒ [tex]R_e=\frac{R_1\times R_2}{R_1 + R_2}[/tex]
⇒ [tex]R_e=\frac{12\times 12}{12 + 12}[/tex]
⇒ [tex]R_e=\frac{144}{24}[/tex]
⇒ [tex]R_e= 6[/tex] Ω
Equivalent resistance of Jason's circuit = [tex]R_1 + R_e =(24+6)=30[/tex] Ω
Circuit 2:
JoAnna's circuit :
According to the question :
The equivalent resistance of both the resistor's are same.
Say the resistance are [tex]R[/tex] and it is equivalent to [tex]R_E[/tex].
And all three resistor are in parallel.
So
⇒ [tex]\frac{1}{R_E} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R}[/tex]
⇒ [tex]\frac{1}{R_E} = \frac{3}{R}[/tex]
⇒ [tex]R=3\times R_E[/tex]
⇒ [tex]R=3\times 30[/tex]
⇒ [tex]R=90[/tex] Ω
The value of JoAnna's resistors = 90 Ω
