Given that R1 =10ohm, R2=40ohm, R3=30ohm, R4= 20ohm and RA is the parallel combination of R1, and R2
whereas RB is the parallel combination of R3
and R4 Combination RA is connected to the positive terminal of 12V battery while
combination RB is connected to the negative
terminal. Ammeter A is connected between the
resistors RA and Rß. a) Find RA and RB, Also
calculate total resistance in the circuit. e) Draw
the above circuit diagram showing combinations connected to battery and Ammeter.
Answers
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Explanation:
Answer: Ra = 8 ohm, Rb = 12 ohm and total resistance is R = 20 ohm
Explanation:
The equivalent resistance of resistors connected in parallel:
\frac{1}{Rp}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...
The equivalent resistance of resistors connected in series is:
Rs=R_1+R_2+R_3+...
It is given that R_1 and R_2 are connected in parallel:
\frac{1}{Ra}=\frac{1}{R_1}+\frac{1}{R_2}\\ \Rightarrow Ra = [ \frac{1}{10}+\frac{1}{40}]^{-1} \\ \Rightarrow Ra = 8 ohm
It is given that R_3 and R_4 are connected in parallel:
\frac{1}{Rb}=\frac{1}{R_3}+\frac{1}{R_4}\\ \Rightarrow Rb = [ \frac{1}{30}+\frac{1}{20}]^{-1} \\ \Rightarrow Ra = 12 ohm
Ra and Rb are connected in series:
Total resistance of the circuit :
R = Ra+ Rb = 8 ohm + 12 ohm = 20 ohm
Answer:
a)
Ra = 40/5 = 8 ohm
Rb = 60/5 = 12 ohm