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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.

Illustration Questions

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?

image
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

Illustration Questions

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, \(V_Q=V_P–IR\)

image

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

VQ = VP–IR

VQ = 30–1×10

VQ = 30–10

VQ = 20 V

image

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.

image
A

15 V

.

B

10 V

C

12 V

D

20 V

Option D 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]

Illustration Questions

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,

\(\Delta V=V_Q–V_P\)

\(\Rightarrow\,\,\,\mathcal{E}=V_Q-V_P\)

\(\Rightarrow\,\,V_Q=V_P +\mathcal{E}\)

image

VQ = VP + e.m.f of battery

VQ = 15 + 10

VQ = 25 V

image

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.

image
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, \(V_Q=V_P-\mathcal{E}–IR\)

 

Illustration Questions

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.

image

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.

\(V_T=V_P+\mathcal E\)

\(V_T=30+50\)

\(V_T=80\ Volt\)

image image

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

image image

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 \(V_P=30\,V\).

image
A

80 V

.

B

90 V

C

70 V

D

60 V

Option B 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)

\(\Delta V=V_P–V_Q\)

\(\Delta V=(V_P–V_C)–(V_Q–V_C)\)

\(\Delta V=\mathcal{E}\,–Ir\)

Illustration Questions

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)

\(V_P–V_Q=(V_P–V_C)–(V_Q–V_C)\)

\(V_P–V_Q=\mathcal{E}\,–IR\)  

image

Potential difference across terminals (P and Q)

\(\Delta V=V_P–V_Q\)

\(\Delta V=\mathcal{E} \,–IR\)

\(\Delta V=10 \,–1×3\)

\(\Delta V=10\,–\,3\)

\(\Delta V=7\,V\)

image

Find the potential difference across the battery.

image
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, \(V_Q=V_P–\mathcal{E}–IR\)

 

Illustration Questions

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)

image image

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

image image

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.

image

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

image
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

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