Informative line

### Emf And Potential

Learn potential difference across resistor & battery, terminal potential difference. Practice equation to calculate potential difference across combination of battery and resistance.

# Potential Difference across a Resistor

• Consider a resistance in which current is flowing, as shown in figure.

• Current always flow in a resistor from point of higher potential to point of lower potential.

## Potential difference across an ideal battery

• An ideal battery B is shown in figure.

• Consider a circuit consisting of two ideal batteries $$B_1$$ and $$B_2$$ as shown in figure

• For B1

Direction of current : P to Q

Potential difference across B1 : VP–VQ    (VP is at positive terminal)

(VQ is at negative terminal)

• For B2

Direction of current : T to S

Potential difference across B2 : VS–VT    (VS is at positive terminal)

(VT is at negative terminal)

• In case of an ideal battery, potential difference across the terminal of battery is independent of direction of current.

#### Which point is at higher potential?

A Point P

B Point Q

C Both at same potential

D None of these

×

Current always flow from higher potential to lower potential through a resistance. Hence, point P is at higher potential.

Thus option  A  is correct.

### Which point is at higher potential?

A

Point P

.

B

Point Q

C

Both at same potential

D

None of these

Option A is Correct

# Potential at any Terminal of Resistance

### Case 1

• Consider a resistor of  R $$\Omega$$ through which current  $$I$$ is flowing.
• Potential at point P = V volt
• Potential at point Q = VP – (voltage drop across resistor 'R')
• VQ = VP – IR

VQ =V – IR

### Case 2

VQ =VP + IR

VQ = V + IR

#### Consider a resistor of resistance $$R=10\ \Omega$$ through which current  $$I=1\ A$$ is passing. If point P is at $$V=30\ V$$, calculate the potential at point Q in the given figure.

A 15 V

B 10 V

C 12 V

D 20 V

×

Since, VP at higher potential and VQ at lower potential.

Potential drop across resistor is given as

$$\Delta V=IR$$

Potential difference across terminal of resistor is given as

$$\Delta V=V_P–V_Q$$

or, $$IR=V_P–V_Q$$

or, $$V_Q=V_P–IR$$

Given: $$V_{P}=30\ V,\ \ \ I=1\ A,\ \ \ R=10\ \Omega$$

VQ = VP–IR

VQ = 30–1×10

VQ = 30–10

VQ = 20 V

### Consider a resistor of resistance $$R=10\ \Omega$$ through which current  $$I=1\ A$$ is passing. If point P is at $$V=30\ V$$, calculate the potential at point Q in the given figure.

A

15 V

.

B

10 V

C

12 V

D

20 V

Option D is Correct

# Potential Difference across Real Battery

• Generally, the resistance of connecting wires is assumed to be zero.
• A practical battery is made of some material which offers resistance to flow of charge within the battery. This resistance is known as internal resistance.

For ideal battery

• Internal resistance, r = 0
• Potential difference = $$\mathcal{E}$$ (e.m.f of battery)

For practical battery

• Internal resistance r$$\neq$$0
• To calculate potential difference of a practical battery, consider a circuit,as shown in figure.

• Current I is flowing in a circuit having battery (practical) with e.m.f  $$\mathcal{E}$$ and internal resistance r.
• Potential difference across terminal (P and Q)

$$\Delta V=V_P–V_Q$$

$$\Delta V=(V_P–V_C)–(V_Q–V_C)$$

$$\Delta V=\mathcal{E}\,–Ir$$

#### Find the potential difference across the battery.

A 16 V

B 14 V

C 7 V

D 6 V

×

Point P is at high potential and point Q at low potential. Taking a point C between battery and internal resistance, potential difference across terminal (P and Q)

$$V_P–V_Q=(V_P–V_C)–(V_Q–V_C)$$

$$V_P–V_Q=\mathcal{E}\,–IR$$

Potential difference across terminals (P and Q)

$$\Delta V=V_P–V_Q$$

$$\Delta V=\mathcal{E} \,–IR$$

$$\Delta V=10 \,–1×3$$

$$\Delta V=10\,–\,3$$

$$\Delta V=7\,V$$

### Find the potential difference across the battery.

A

16 V

.

B

14 V

C

7 V

D

6 V

Option C is Correct

# Potential Difference across Combination of Two Batteries and Resistances

### Case 1

• Consider a circuit in which a battery with e.m.f is connected with load resistance R, through which current is flowing in the circuit.
• Potential at point P is VP.

• To calculate potential at point Q, mark a point T between battery and resistance.

• Divide the circuit into two parts at T and then calculate separately.

So,  $$V_Q=V_P+\mathcal{E}–IR$$

### Case 2

• Consider a circuit in which a battery with e.m.f is connected with load resistance R, through which current is flowing in the circuit.
• Potential at point P is VP.

• To calculate potential at point Q, mark a point T between battery and resistance.

• Divide the circuit into two parts at T and then calculate separately.

So, $$V_Q=V_P–\mathcal{E}–IR$$

#### Two real batteries are connected,as shown in figure. If VA = V Volt, choose the incorrect option.

A $$V_C=V–[(\mathcal{E}_1+\mathcal{E}_2)+I(r_1+R_1+r_2+R_2)]$$

B $$V_C=V–[(\mathcal{E}_1-\mathcal{E}_2)+I(r_1+R_1-r_2+R_2)]$$

C $$V_B=V_A–\mathcal{E}_1-I(r_1+R_1)$$

D $$V_C=V_B–\mathcal{E}_2-I(r_2+R_2)$$

×

Break circuit from point B into parts and calculate separately.

$$V_B=V_A–\mathcal{E}_1-I(r_1+R_1)$$  ...(i)

$$V_C=V_B–\mathcal{E}_2-I(r_2+R_2)$$  ...(ii)

From (i) and (ii)

$$V_C=V_A–\mathcal{E}_1-I(r_1+R_1)–\mathcal{E}_2-I(r_2+R_2)$$

$$V_C=V–[(\mathcal{E}_1+\mathcal{E}_2)+I(r_1+R_1+r_2+R_2)]$$

Hence,option A, C, D are correct and option B is incorrect.

### Two real batteries are connected,as shown in figure. If VA = V Volt, choose the incorrect option.

A

$$V_C=V–[(\mathcal{E}_1+\mathcal{E}_2)+I(r_1+R_1+r_2+R_2)]$$

.

B

$$V_C=V–[(\mathcal{E}_1-\mathcal{E}_2)+I(r_1+R_1-r_2+R_2)]$$

C

$$V_B=V_A–\mathcal{E}_1-I(r_1+R_1)$$

D

$$V_C=V_B–\mathcal{E}_2-I(r_2+R_2)$$

Option B is Correct

# Potential Difference across an Ideal Battery

• Consider a battery with e.m.f  $$\mathcal{E}$$
• For an ideal battery, potential difference across terminal is independent of direction of flow of current.

### Case 1

Potential at point P = V volt

Potential at point Q,

VQ = V– e.m.f of the battery

$$V_{Q}=V-\mathcal E$$

### Case 2

VQ = V– e.m.f. of battery

$$V_{Q}=V-\mathcal E$$

[Potential at point is independent of direction of current]

### Case 3

VQ = VP+ e.m.f of battery

$$V_{Q}=V+\mathcal E$$

### Case 4

VQ =  e.m.f of battery + VP

$$V_Q=\mathcal E+V$$

[Potential at point across battery is independent of direction of current]

#### A battery with e.m.f $$(\mathcal{E}=10\ V)$$ is connected in a circuit and current of  $$I=1\,A$$ is flowing through it. If potential at P is $$V_P=15\,V$$, calculate potential at Q.

A 20 V

B 15 V

C 25 V

D 10 V

×

Since, point Q is at high potential and point P is at low potential. So, potential drop across battery,

$$\Delta V=V_Q–V_P$$

$$\Rightarrow\,\,\,\mathcal{E}=V_Q-V_P$$

$$\Rightarrow\,\,V_Q=V_P +\mathcal{E}$$

VQ = VP + e.m.f of battery

VQ = 15 + 10

VQ = 25 V

### A battery with e.m.f $$(\mathcal{E}=10\ V)$$ is connected in a circuit and current of  $$I=1\,A$$ is flowing through it. If potential at P is $$V_P=15\,V$$, calculate potential at Q.

A

20 V

.

B

15 V

C

25 V

D

10 V

Option C is Correct

# Potential Difference across Combination of Battery and Resistance

### Case 1

• Consider a circuit in which a battery with e.m.f is connected with load resistance R, through which current is flowing in the circuit.
• Potential at point P is VP.

• To calculate potential at point Q, mark a point T between battery and resistance.

• Divide the circuit into two parts at T and then calculate separately.

So,

$$V_Q=V_P+\mathcal{E}–IR$$

### Case 2

• Consider a circuit in which a battery with e.m.f is connected with load resistance R, through which current is flowing in the circuit.
• Potential at point P is VP.

• To calculate potential at point Q, mark a point T between battery and resistance.

• Divide the circuit into two parts at T and then calculate separately.

So, $$V_Q=V_P-\mathcal{E}–IR$$

#### A battery is connected in a circuit with load resistance $$R=10\,\Omega$$ ,through which $$I=1\,A$$ current is flowing. Calculate potential at point Q if potential at point P is at $$V_P=30\,V$$.

A 80 V

B 90 V

C 70 V

D 60 V

×

To calculate potential at point Q, mark a point T between battery and resistance.

Break the circuit and calculate separately.

While traversing from P to T the e.m.f of battery$$(\mathcal{E})$$ is 50 V and T is at high potential.

$$V_T=V_P+\mathcal E$$

$$V_T=30+50$$

$$V_T=80\ Volt$$

While traversing from T to Q the potential drop across resistor is $$IR$$

VQ = V+ IR

VQ = 80 + 1×10

VQ = 80+10

VQ = 90 V

### A battery is connected in a circuit with load resistance $$R=10\,\Omega$$ ,through which $$I=1\,A$$ current is flowing. Calculate potential at point Q if potential at point P is at $$V_P=30\,V$$.

A

80 V

.

B

90 V

C

70 V

D

60 V

Option B is Correct