If a 12V battery is connected to the arrangement of resistances shown in figure
calculte (i) the total effective resistance of arrangement and (ii) current flowing
in the circuit.
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R = 15 Ω
I = 0.8 Ampere
Step-by-step explanation:
From the given figure, we can say that ;
R1 and R2 are in the series, for series we add resistance simply as shown below,
R1 + R2 = ( 10 + 20 ) Ω
⇒ R1 + R2 = 30 Ω
And, Resistance R3 and R4 are in the series, therefore
R3 + R4 = ( 5 + 25 ) Ω
⇒ R3 + R4 = 30 Ω
Now, R1 + R2 and R3 + R4 are parallel resistors,
⇒ 1 / 30 + 1 / 30
⇒ 1 + 1 / 30
⇒ 2 / 30
⇒ 1 / 15
Hence, total resistance = 15 Ω
We know that,
V = IR
[ ∵ where, V = volt, I = current, R = resistance]
⇒ 12 = I * 15
⇒ I = 12 / 15
⇒ I = 0.8 Ampere
Hence, total resistance is 15 Ω and total current flowing is 0.8 Ampere.
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