A circuit has three resistors connected to a 3.00-V battery in series: R_1 = 15.
ID: 2081133 • Letter: A
Question
A circuit has three resistors connected to a 3.00-V battery in series: R_1 = 15.0 ohm, R_2 = 10.0 ohm, and R_3 = 25.0 ohm. Find the equivalent resistance, R_eq, the total current in the circuit, I_tot, and the voltage across EACH resistor, V_1, V_2, and V_3. The same resistors are now connected to the 3.00-V battery in parallel. Find the equivalent resistance R_eq, the total current in the circuit, I_tot, and the current through EACH resistor, I_1, I_2, and I_3. A fourth resistor, R_4 = 13.0 ohm, is now added, and the four resistors are arranged according to the diagram above. Find the equivalent resistance, R_eq, the total current in the circuit, I_tot, the current through EACH resistor, I_1, I_2, I_3, and I_4 and the voltage across EACH resistor, V_1, V_2, V_3 and V_4. A fourth resistor, R_4 = 13.0 ohm, is now added, and the four resistors are arranged according to the diagram above. Find the equivalent resistance, R_eq, the total current in the circuit, I_tot, the current through EACH resistor, I_1, I_2, I_3, and I_4 and the voltage across EACH resistor, V_1, V_2, V_3 and V_4.Explanation / Answer
1) V=3.00 Volts
R1= 15 Ohms
R2=10 Ohms
R3=25 Ohms
since resistors are in series i.e Requialent =R1+R2+R3
= 15+10+25
=50 Ohms
Total current(I)= V/Requialent
=3/50
=0.06 Amperes
V1=iR1
=0.06*15
=0.9 Volts
V2= iR2
=0.06*10
=0.6 Volts
V3=iR3
=0.06*25
=1.5 Volts
......................................................................................................................................................................
2) since resistors are connected in parallel voltage is same i.e V1= V2=V3
Hence i1= V/R1
=3/15
=0.5 aAmperes
i2=V/R2
=3/10
=0.3 Amperes
i3=V/R3
=3/25
=0.12 Amperes
Total current(I total ) =i1+i2+i3
=0.5+0.3+0.12
=0.92
......................................................................................................................................................................
3) since R2//R3 are parallel
R total =(R2+R3)/(R2R3)
=(10+25)/(250)
=0.14 ohms
now R total ,R1 and R4 are in series
Requialent =R1+Rtotal+R4
=15+0.14+13
=28.14
i1 =V/R1
=3/15
=0.2 Amperes
i2= V/R2
=3/10
=0.3 Amperes
i3=V/R3
=3/25
=0.12 Ampers
i4=V/R3
=3/13
=0.23 Amperes
Total current =i1+i2+i3+i4
=0.2+0.3+0.12+0.23
=0.85 Amperes
V1 =i1*R1
=0.2*15
=3 Volts
V2= I2*R2
=0.3*10
=3 Volts
V3=i3*R3
=0.12*25
=3 volts
V4=i4*R4
=0.23*13
=2.99 Volts
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