Potential dividers

A potential divider consists of a battery of e.m.f. 6.00 V and negligible internal resistance connected in series with a resistor of resistance 120 Ω and a variable resistor of resistance 0 → 200 Ω. Determine the range of potential difference that can be obtained across the fixed resistor.

When the variable resistor is at 0Ω: `V = (120 / (120 + 0)) xx6 = "6 V"`

When the variable resistor is at 200Ω: `V = (120 / (120 + 200)) xx 6 = (120 / 320) xx 6 = "2.25 V"`

The range is 2.25 V → 6.00 V

A battery of electromotive force (e.m.f.) 9.0 V and internal resistance 1.0 Ω is connected to a fixed resistor of resistance 5.0 Ω and a potentiometer of maximum resistance 3.0 Ω, as shown.
The sliding contact of the potentiometer is moved over its full range of movement.

What is the maximum value of the potential difference that is measured by the voltmeter?

A. 3.0 V.

B. 3.4 V.

C. 4.5 V.

D. 5.4 V.

Answer: A

Voltmeter measures the voltage across the potentiometer, which varies from 0 V to the maximum voltage drop across it.

Total resistance in the circuit (when potentiometer at max resistance):

`R_("total") =r+R+R_("pot")= 1 + 5 + 3 = 9 " " Omega`

Total current in the circuit: `I = E / R_("total") = 9 / 9 = "1 A"`

Voltage drop across the potentiometer (maximum case):

`V_("pot") = I xx R_("pot") = 1 xx 3 = "3 V"`

A battery of e.m.f. 6.0V and negligible internal resistance is connected to a network of resistors and a voltmeter, as shown in figure. Resistor Y has a resistance of 24 Ω and resistor Z has a resistance of 32 Ω.

The resistance R_X​ of the variable resistor X is adjusted until the voltmeter reads 4.8 V. Calculate:
a. The current in resistor Z.

b. The total power provided by the battery.

c. The number of conduction electrons that move through the battery in a time interval of 25 s.

d. The total resistance of X and Y connected in parallel.

e. The resistance R_X​.

a. `I_Z = 4.8 / 32 = "0.15 A"`

b. `P = VxxI= 6 xx 0.15 = "0.90 W"`

c. Total charge: `Q = Ixxt= 0.15 xx 25 = "3.75 C"`

Each electron carries e = 1.6×10⁻¹⁹ C, so number of electrons: 

`n =Q/e= 3.75 / (1.6 xx 10^-19) ≈ 2.34 xx 10^19`

d. The remaining voltage drop after Z: `V _("XY") ​ =6.0−4.8="1.2 V"`

Total resistance of X‖Y: `R_("XY") = 1.2 / 0.15 = 8 " " Omega`

e. R_Y​ = 24 Ω.

Total parallel resistance R_P = 8 Ω.

`1/R_P = 1/R_X + 1/24 => 1/8 = 1/R_X + 1/24 => R_X = 12 " " Omega`

Potential differences across two resistors of resistances R₁ and R₂ are compared using a potentiometer wire (uniform resistance wire) in the electrical circuit shown.

One terminal of a galvanometer is connected to point X. The galvanometer reads zero when its
other terminal is connected to a point that is a distance of 60 cm from one end of the
potentiometer wire.

One terminal of a second galvanometer is connected to point Y. This galvanometer reads zero
when its other terminal is connected to a point that is a distance of 80 cm from the same end of
the potentiometer wire.

What is the ratio `R_2/R_1` ?

R₂​ corresponds to 80 cm from the same end, so the length representing only R₂:

80 cm − 60 cm = 20 cm.

Hence, V₁​ ∝ 60 and V₂ ∝ 20.

Since both resistors are connected in series with the same current I:

`V_1 = I xx R_1, " " V_2 = I xx R_2 => V_2 / V_1 = R_2 / R_1`

`R_2 / R_1 = V_2 / V_1 = 20 / 60 = 1 / 3`

A circuit consists of a battery, a high-resistance voltmeter and four fixed resistors, as shown. The battery has an electromotive force (e.m.f.) of 15.0 V and negligible internal resistance.
What is the reading on the voltmeter?

Left side: 24 Ω (top) and 6 Ω (bottom)

Voltage at the midpoint (between 24 Ω and 6 Ω), relative to the bottom (0 V):

`V_("left") = (6 / (24 + 6)) xx 15 = (6 / 30) xx 15 = "3 V"`

Right side: 6 Ω (top) and 9 Ω (bottom) 

Voltage at the midpoint: 

`V_("right") = (9 / (6 + 9)) xx 15 = (9 / 15) xx 15 = 9 V`

Voltage across the voltmeter (difference between midpoints):

`V = V_("right") - V_("left") = 9 - 3 = "6 V"`

Three batteries and three identical resistors are connected in a circuit PQR, as shown.
The batteries have negligible internal resistance.

What is the potential difference between points P and Q?

A. 1.5 V. 

B. 2.1 V. 

C. 7.1 V. 

D. 12.1 V.

Answer: B

Let’s the potential at point R is zero.

Moving from R to P through the 8.4 V battery and resistor: 

Voltage at point P: `V_P = 8.4 - IxxR`

Moving from R to Q through the 6.3 V battery and resistor:

Voltage at point Q: `V_Q = 6.3 - IxxR`

Find V_P − V_Q​: `V_P - V_Q = (8.4 - IxxR) - (6.3 - IxxR) = "2.1 V"`

In the circuit shown, the 6.0 V battery has negligible internal resistance. Resistors R₁​ and R₂ and the voltmeter each have a resistance of 100 kΩ.
What is the current in the resistor R₂?

A. 20 μA. 

B. 30 μA.

C. 40 μA. 

D. 60 μA.

Answer: C

Since both R₂​ and the voltmeter are 100 kΩ and are in parallel, the equivalent resistance:

`R_("eq") = 1 / (1/100000 + 1/100000) = 50000 " " Omega`

Total circuit resistance: `R_("total") = R_1+R_("eq")= 100000 + 50000 = 150000 " " Omega`

Total current from the battery: `I_("total") = V/R= 6 / 150000 = 40 " " mu"A"`

Current through R₂: `I_(R_2) =I_("total")/2 = 40 / 2 = 20 " " mu"A"`

A potential divider consists of two resistors of resistances R₁​ and R_2​ connected in series across a source of potential difference (p.d.) V_in. The p.d. across R_1​ is V_out.
Which changes to R₁ and R₂ will increase the value of V_out?

Answer: B

The formula for the output voltage across R₁: `V_"out" = (R_1 / (R_1 + R_2)) xx V_"in"`

To increase V_out​, we need the fraction `(R_1 / (R_1 + R_2))`​​ to increase.

That happens when: R₁​ increases (numerator increases) and/or R₂ decreases (denominator decreases).

A. Correct: `V_"out" = (2R_1) / (2R_1 + 2R_2) = R_1 / (R_1 + R_2)`

No change to the voltage ratio.

B. Correct:

R₁​ is doubled → increases numerator.

R₂ is halved → decreases denominator.

Fraction increases ⇒ V_out​ increases.

C. Incorrect: Numerator decreases, denominator increases ⇒ V_out​ decreases.

D. Incorrect: `V_"out" = (0.5R_1) / (0.5R_1 + 0.5R_2) = R_1 / (R_1 + R_2)`

No change to the voltage ratio.

A power supply and a solar cell are compared using the potentiometer circuit shown.
The potentiometer wire PQ is 100.0 cm long and has a resistance of 5.00 Ω. The power supply has an e.m.f. of 2.000 V and the solar cell has an e.m.f. of 5.00 mV.

Which resistance R must be used so that the galvanometer reads zero when PS = 40.0 cm?

A. 395 Ω.

B. 405 Ω.

C. 795 Ω.

D. 805 Ω.

Answer: C

Total resistance in the circuit: R + 5.00 Ω

Current from the 2.000 V power supply: `I = 2.000 / (R + 5)`

Voltage per cm: `Ixx5.00/100=Ixx0.05`

So, voltage across 40.0 cm: `V_("PS") = I xx (40.00xx0.05) = I xx 2.0`

`V_("PS") = (2 / (R + 5)) xx2 = 4 / (R + 5)`

At null point (galvanometer reads zero), this voltage must equal the solar cell e.m.f:

`4 / (R + 5) = 0.005`

`R = 800 - 5 = 795 " "Omega`

A potential divider circuit is designed to detect the difference in temperature between two different places.
The cell has electromotive force (e.m.f.) 20 mV and negligible internal resistance.

Initially, thermistors X and Y are at the same temperature and have the same resistance. The voltmeter reads 10 mV. X is then placed in a cold environment and its resistance doubles. Y is placed in a warm environment and its resistance halves.

What is the new reading on the voltmeter?

Answer: A

Initially:

R_X = R

R_Y = R
Total resistance = R + R = 2R

Using voltage division rule: `V = (R / (R + R)) xx 20 = (1/2) xx 20 = "10 mV"`

After temperature change:

R_X = 2R

R_Y = R/2

Total resistance = 2R + R/2 = 5R/2

New voltmeter reading:

`V = (R_Y) / (R_X + R_Y) xx V_("total") = (R/2) / (2R + R/2) xx 20 = (R/2) / (5R/2) xx 20 = (1/5) xx 20 = "4 mV"`​