In a series connection, R1= 5 ohm, R2= 15 ohm, R3 = 10 ohm and is connected to a battery 5V. Calculate the potential difference at R2.
Answers
Explanation:
equivalent resistance is 30
V=IR
I=V/R
=5/30
=1/6
potential difference at R2
V = 1/6×15
=2.5V
Answer :
- The potential difference at the resistor (R2) is 2.5 V.
Explanation :
Given :
- Resistance in the resistor, R1 = 5 Ω
- Resistance in the resistor, R2 = 15 Ω
- Resistance in the resistor, R3 = 10 Ω
- Voltage of the circuit = 5 V
To find :
- Potential difference at the resistor , R2 = ?
Knowledge required :
In a series circuit ,
- The current flowing through the Resistors will be equal,i.e. If the current I flows through the circuit R1 , the same current I will also flow through the circuit R2.
- The Potential Difference (voltage) in a series circuit are different at different circuits.i.e, if the voltage V1 is flowing through the circuit R1 , then the voltage V1 will flow through the circuit R2 (V1 ≠ V2).
Now ,
Formulae and law , required for the solution :
- Total resistance in a series circuit :
⠀⠀⠀⠀⠀⠀⠀⠀⠀Re = R1 + R2 + R3 + ... + Rn
Where :
- Re = Equivalent resistance
- R = Resistance the circuit
- Ohm's law :
⠀⠀⠀⠀⠀⠀⠀⠀⠀V = IR
Where :
- V = Potential Difference
- I = Current
- R = Resistance
Solution :
Equivalent resistance in the circuit :
Using the formula for total resistance in the circuit and substituting the values in it, we get :
==> Re = R1 + R2 + R3
==> Re = 5 + 15 + 10
==> Re = 30
∴ Re = 30 Ω
Hence the equivalent resistance in the circuit is 30 Ω.
Current in the circuit :
Using the ohm's law and substituting the values in it, we get :
==> V = IR
==> 5 = I × 30
==> 5/30 = I
==> 0.166(approx.) = I
==> 0.17 = I
∴ I = 0.17 A.
Hence the current flowing in the circuit is 0.17 A.
Potential difference at R2 :
Using the ohm's law and substituting the values in it, we get :
==> V = IR
==> V = 0.17 × 15
==> V = 2.505
==> V = 2.5 (approx.)
∴ V = 2.5 V
Therefore,
- The potential difference in the circuit R2 is 2.5 V.