Learn potential difference across resistor & battery, terminal potential difference. Practice equation to calculate potential difference across combination of battery and resistance.
Direction of current : P to Q
Potential difference across B1 : VP–VQ (VP is at positive terminal)
(VQ is at negative terminal)
Direction of current : T to S
Potential difference across B2 : VS–VT (VS is at positive terminal)
(VT is at negative terminal)
VQ = VP – IR
VQ =V – IR
VQ =VP + IR
VQ = V + IR
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
Option D is Correct
For ideal battery
For practical battery
\(\Delta V=V_P–V_Q\)
\(\Delta V=(V_P–V_C)–(V_Q–V_C)\)
\(\Delta V=\mathcal{E}\,–Ir\)
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)
\(\Delta V=V_P–V_Q\)
\(\Delta V=\mathcal{E} \,–IR\)
\(\Delta V=10 \,–1×3\)
\(\Delta V=10\,–\,3\)
\(\Delta V=7\,V\)
Option C is Correct
So, \(V_Q=V_P+\mathcal{E}–IR\)
So, \(V_Q=V_P–\mathcal{E}–IR\)
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.
Option B is Correct
Potential at point P = V volt
Potential at point Q,
VQ = VP – e.m.f of the battery
\(V_{Q}=V-\mathcal E\)
VQ = VP – e.m.f. of battery
\(V_{Q}=V-\mathcal E\)
[Potential at point is independent of direction of current]
VQ = VP+ e.m.f of battery
\(V_{Q}=V+\mathcal E\)
VQ = e.m.f of battery + VP
\(V_Q=\mathcal E+V\)
[Potential at point across battery is independent of direction of current]
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
Option C is Correct
So,
\(V_Q=V_P+\mathcal{E}–IR\)
So, \(V_Q=V_P-\mathcal{E}–IR\)
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 = VT + IR
VQ = 80 + 1×10
VQ = 80+10
VQ = 90 V
Option B is Correct
