152677
You are given resistance wire of length $50 \mathrm{~cm}$ and a battery of negligible resistance. In which of the following cases is largest amount of heat generated?
1 When the wire is connected to the battery directly
2 When the wire is divided into two parts and both the parts are connected to the battery in parallel.
3 When the wire is divided into four parts and all the four parts are connected to the battery in parallel.
4 When only half of the wire is connected to the battery.
Explanation:
C Given, Length of wire $(l)=50 \mathrm{~cm}$ Formula for heating effect of current is $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\because \mathrm{H} \propto \frac{\mathrm{V}^{2}}{\mathrm{R}}$ According to options- Option (a): When the wire is connected to the battery directly If connected directly, then resistance is $\mathrm{R}$. Option (b): When the wire is divided into two parts and both the parts are connected to the battery in parallel. If the wire is divided into two equal parts, then resistance becomes $\frac{R}{4}$ Option (c): When the wire is divided into four parts and all the four parts are connected to the battery in parallel. If the wire is divided into four equal parts, then resistance becomes $\frac{\mathrm{R}}{16}$ Hence, to get the largest amount of heat generated in 50 $\mathrm{cm}$ long wire it has to be cut in equal four parts are connected in parallel to the battery since, in parallel combination voltage across each part will be same.
JIPMER-2017
Current Electricity
152678
Three identical resistors each of resistance $R$ are connected to an ideal cell of voltage $V$ as shown. Total power dissipated in all three resistors is
1 $\frac{3 V^{2}}{2 R}$
2 $\frac{3 V^{2}}{R}$
3 $\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$
4 $\frac{2 \mathrm{~V}^{2}}{3 \mathrm{R}}$
Explanation:
D From the circuit diagram, Equivalent resistance of circuit, $R_{\text {eq }}=R+\frac{R \times R}{R+R}$ $R_{\text {eq }}=R+\frac{R^{2}}{2 R}$ $R_{\text {eq }}=\frac{2 R^{2}+R^{2}}{2 R}=\frac{3 R^{2}}{2 R}=\frac{3 R}{2}$ We know that, $\text { Power }(\mathrm{P}) =\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{\mathrm{V}^{2}}{\frac{3 \mathrm{R}}{2}}$ $\mathrm{P} =\frac{2 \mathrm{~V}^{2}}{3 \mathrm{R}}$
Current Electricity
152679
In the given circuit, cell $E$ has internal resistance of $r=2 \Omega$. What is the value of resistance $R$ so that power delivered to resistor $R$ is maximum?
1 $2 \Omega$
2 $3 \Omega$
3 $5 \Omega$
4 $1 \Omega$
Explanation:
C Since, equivalent internal resistance of equivalent cell across the external resistor $\mathrm{R}$ is $2+3=5 \Omega$. Hence power delivered to $\mathrm{R}$ will be maximum if $\mathrm{R}=$ $5 \Omega$ (according to maximum power transfer theorem).
UPSEE - 2017
Current Electricity
152680
A30V-90W lamp is operated on a $120 \mathrm{~V}$ DC line. A resistor is connected in series with the lamp in order to glow it properly. The value of resistance is
1 $10 \Omega$
2 $30 \Omega$
3 $20 \Omega$
4 $40 \Omega$
Explanation:
B Given, Voltage $(\mathrm{V})=30$ Volt Power $(\mathrm{P})=90$ Watt We know that, Power $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}_{0}}$ $\mathrm{R}_{0}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}_{0}=\frac{30^{2}}{90}$ $\mathrm{R}_{0}=\frac{900}{90}=10 \Omega$ Therefore, current in the lamp $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}_{0}}=\frac{30}{10}=3 \mathrm{~A}$ Now, the lamp is operated on $120 \mathrm{~V} \mathrm{DC}$, Applying KVL in loop, $120-3 \times 10-3 \mathrm{R}=0$ $3 \mathrm{R}=90$ $\mathrm{R}=30 \Omega$
152677
You are given resistance wire of length $50 \mathrm{~cm}$ and a battery of negligible resistance. In which of the following cases is largest amount of heat generated?
1 When the wire is connected to the battery directly
2 When the wire is divided into two parts and both the parts are connected to the battery in parallel.
3 When the wire is divided into four parts and all the four parts are connected to the battery in parallel.
4 When only half of the wire is connected to the battery.
Explanation:
C Given, Length of wire $(l)=50 \mathrm{~cm}$ Formula for heating effect of current is $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\because \mathrm{H} \propto \frac{\mathrm{V}^{2}}{\mathrm{R}}$ According to options- Option (a): When the wire is connected to the battery directly If connected directly, then resistance is $\mathrm{R}$. Option (b): When the wire is divided into two parts and both the parts are connected to the battery in parallel. If the wire is divided into two equal parts, then resistance becomes $\frac{R}{4}$ Option (c): When the wire is divided into four parts and all the four parts are connected to the battery in parallel. If the wire is divided into four equal parts, then resistance becomes $\frac{\mathrm{R}}{16}$ Hence, to get the largest amount of heat generated in 50 $\mathrm{cm}$ long wire it has to be cut in equal four parts are connected in parallel to the battery since, in parallel combination voltage across each part will be same.
JIPMER-2017
Current Electricity
152678
Three identical resistors each of resistance $R$ are connected to an ideal cell of voltage $V$ as shown. Total power dissipated in all three resistors is
1 $\frac{3 V^{2}}{2 R}$
2 $\frac{3 V^{2}}{R}$
3 $\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$
4 $\frac{2 \mathrm{~V}^{2}}{3 \mathrm{R}}$
Explanation:
D From the circuit diagram, Equivalent resistance of circuit, $R_{\text {eq }}=R+\frac{R \times R}{R+R}$ $R_{\text {eq }}=R+\frac{R^{2}}{2 R}$ $R_{\text {eq }}=\frac{2 R^{2}+R^{2}}{2 R}=\frac{3 R^{2}}{2 R}=\frac{3 R}{2}$ We know that, $\text { Power }(\mathrm{P}) =\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{\mathrm{V}^{2}}{\frac{3 \mathrm{R}}{2}}$ $\mathrm{P} =\frac{2 \mathrm{~V}^{2}}{3 \mathrm{R}}$
Current Electricity
152679
In the given circuit, cell $E$ has internal resistance of $r=2 \Omega$. What is the value of resistance $R$ so that power delivered to resistor $R$ is maximum?
1 $2 \Omega$
2 $3 \Omega$
3 $5 \Omega$
4 $1 \Omega$
Explanation:
C Since, equivalent internal resistance of equivalent cell across the external resistor $\mathrm{R}$ is $2+3=5 \Omega$. Hence power delivered to $\mathrm{R}$ will be maximum if $\mathrm{R}=$ $5 \Omega$ (according to maximum power transfer theorem).
UPSEE - 2017
Current Electricity
152680
A30V-90W lamp is operated on a $120 \mathrm{~V}$ DC line. A resistor is connected in series with the lamp in order to glow it properly. The value of resistance is
1 $10 \Omega$
2 $30 \Omega$
3 $20 \Omega$
4 $40 \Omega$
Explanation:
B Given, Voltage $(\mathrm{V})=30$ Volt Power $(\mathrm{P})=90$ Watt We know that, Power $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}_{0}}$ $\mathrm{R}_{0}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}_{0}=\frac{30^{2}}{90}$ $\mathrm{R}_{0}=\frac{900}{90}=10 \Omega$ Therefore, current in the lamp $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}_{0}}=\frac{30}{10}=3 \mathrm{~A}$ Now, the lamp is operated on $120 \mathrm{~V} \mathrm{DC}$, Applying KVL in loop, $120-3 \times 10-3 \mathrm{R}=0$ $3 \mathrm{R}=90$ $\mathrm{R}=30 \Omega$
152677
You are given resistance wire of length $50 \mathrm{~cm}$ and a battery of negligible resistance. In which of the following cases is largest amount of heat generated?
1 When the wire is connected to the battery directly
2 When the wire is divided into two parts and both the parts are connected to the battery in parallel.
3 When the wire is divided into four parts and all the four parts are connected to the battery in parallel.
4 When only half of the wire is connected to the battery.
Explanation:
C Given, Length of wire $(l)=50 \mathrm{~cm}$ Formula for heating effect of current is $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\because \mathrm{H} \propto \frac{\mathrm{V}^{2}}{\mathrm{R}}$ According to options- Option (a): When the wire is connected to the battery directly If connected directly, then resistance is $\mathrm{R}$. Option (b): When the wire is divided into two parts and both the parts are connected to the battery in parallel. If the wire is divided into two equal parts, then resistance becomes $\frac{R}{4}$ Option (c): When the wire is divided into four parts and all the four parts are connected to the battery in parallel. If the wire is divided into four equal parts, then resistance becomes $\frac{\mathrm{R}}{16}$ Hence, to get the largest amount of heat generated in 50 $\mathrm{cm}$ long wire it has to be cut in equal four parts are connected in parallel to the battery since, in parallel combination voltage across each part will be same.
JIPMER-2017
Current Electricity
152678
Three identical resistors each of resistance $R$ are connected to an ideal cell of voltage $V$ as shown. Total power dissipated in all three resistors is
1 $\frac{3 V^{2}}{2 R}$
2 $\frac{3 V^{2}}{R}$
3 $\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$
4 $\frac{2 \mathrm{~V}^{2}}{3 \mathrm{R}}$
Explanation:
D From the circuit diagram, Equivalent resistance of circuit, $R_{\text {eq }}=R+\frac{R \times R}{R+R}$ $R_{\text {eq }}=R+\frac{R^{2}}{2 R}$ $R_{\text {eq }}=\frac{2 R^{2}+R^{2}}{2 R}=\frac{3 R^{2}}{2 R}=\frac{3 R}{2}$ We know that, $\text { Power }(\mathrm{P}) =\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{\mathrm{V}^{2}}{\frac{3 \mathrm{R}}{2}}$ $\mathrm{P} =\frac{2 \mathrm{~V}^{2}}{3 \mathrm{R}}$
Current Electricity
152679
In the given circuit, cell $E$ has internal resistance of $r=2 \Omega$. What is the value of resistance $R$ so that power delivered to resistor $R$ is maximum?
1 $2 \Omega$
2 $3 \Omega$
3 $5 \Omega$
4 $1 \Omega$
Explanation:
C Since, equivalent internal resistance of equivalent cell across the external resistor $\mathrm{R}$ is $2+3=5 \Omega$. Hence power delivered to $\mathrm{R}$ will be maximum if $\mathrm{R}=$ $5 \Omega$ (according to maximum power transfer theorem).
UPSEE - 2017
Current Electricity
152680
A30V-90W lamp is operated on a $120 \mathrm{~V}$ DC line. A resistor is connected in series with the lamp in order to glow it properly. The value of resistance is
1 $10 \Omega$
2 $30 \Omega$
3 $20 \Omega$
4 $40 \Omega$
Explanation:
B Given, Voltage $(\mathrm{V})=30$ Volt Power $(\mathrm{P})=90$ Watt We know that, Power $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}_{0}}$ $\mathrm{R}_{0}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}_{0}=\frac{30^{2}}{90}$ $\mathrm{R}_{0}=\frac{900}{90}=10 \Omega$ Therefore, current in the lamp $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}_{0}}=\frac{30}{10}=3 \mathrm{~A}$ Now, the lamp is operated on $120 \mathrm{~V} \mathrm{DC}$, Applying KVL in loop, $120-3 \times 10-3 \mathrm{R}=0$ $3 \mathrm{R}=90$ $\mathrm{R}=30 \Omega$
152677
You are given resistance wire of length $50 \mathrm{~cm}$ and a battery of negligible resistance. In which of the following cases is largest amount of heat generated?
1 When the wire is connected to the battery directly
2 When the wire is divided into two parts and both the parts are connected to the battery in parallel.
3 When the wire is divided into four parts and all the four parts are connected to the battery in parallel.
4 When only half of the wire is connected to the battery.
Explanation:
C Given, Length of wire $(l)=50 \mathrm{~cm}$ Formula for heating effect of current is $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\because \mathrm{H} \propto \frac{\mathrm{V}^{2}}{\mathrm{R}}$ According to options- Option (a): When the wire is connected to the battery directly If connected directly, then resistance is $\mathrm{R}$. Option (b): When the wire is divided into two parts and both the parts are connected to the battery in parallel. If the wire is divided into two equal parts, then resistance becomes $\frac{R}{4}$ Option (c): When the wire is divided into four parts and all the four parts are connected to the battery in parallel. If the wire is divided into four equal parts, then resistance becomes $\frac{\mathrm{R}}{16}$ Hence, to get the largest amount of heat generated in 50 $\mathrm{cm}$ long wire it has to be cut in equal four parts are connected in parallel to the battery since, in parallel combination voltage across each part will be same.
JIPMER-2017
Current Electricity
152678
Three identical resistors each of resistance $R$ are connected to an ideal cell of voltage $V$ as shown. Total power dissipated in all three resistors is
1 $\frac{3 V^{2}}{2 R}$
2 $\frac{3 V^{2}}{R}$
3 $\frac{\mathrm{V}^{2}}{3 \mathrm{R}}$
4 $\frac{2 \mathrm{~V}^{2}}{3 \mathrm{R}}$
Explanation:
D From the circuit diagram, Equivalent resistance of circuit, $R_{\text {eq }}=R+\frac{R \times R}{R+R}$ $R_{\text {eq }}=R+\frac{R^{2}}{2 R}$ $R_{\text {eq }}=\frac{2 R^{2}+R^{2}}{2 R}=\frac{3 R^{2}}{2 R}=\frac{3 R}{2}$ We know that, $\text { Power }(\mathrm{P}) =\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}=\frac{\mathrm{V}^{2}}{\frac{3 \mathrm{R}}{2}}$ $\mathrm{P} =\frac{2 \mathrm{~V}^{2}}{3 \mathrm{R}}$
Current Electricity
152679
In the given circuit, cell $E$ has internal resistance of $r=2 \Omega$. What is the value of resistance $R$ so that power delivered to resistor $R$ is maximum?
1 $2 \Omega$
2 $3 \Omega$
3 $5 \Omega$
4 $1 \Omega$
Explanation:
C Since, equivalent internal resistance of equivalent cell across the external resistor $\mathrm{R}$ is $2+3=5 \Omega$. Hence power delivered to $\mathrm{R}$ will be maximum if $\mathrm{R}=$ $5 \Omega$ (according to maximum power transfer theorem).
UPSEE - 2017
Current Electricity
152680
A30V-90W lamp is operated on a $120 \mathrm{~V}$ DC line. A resistor is connected in series with the lamp in order to glow it properly. The value of resistance is
1 $10 \Omega$
2 $30 \Omega$
3 $20 \Omega$
4 $40 \Omega$
Explanation:
B Given, Voltage $(\mathrm{V})=30$ Volt Power $(\mathrm{P})=90$ Watt We know that, Power $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}_{0}}$ $\mathrm{R}_{0}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}_{0}=\frac{30^{2}}{90}$ $\mathrm{R}_{0}=\frac{900}{90}=10 \Omega$ Therefore, current in the lamp $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}_{0}}=\frac{30}{10}=3 \mathrm{~A}$ Now, the lamp is operated on $120 \mathrm{~V} \mathrm{DC}$, Applying KVL in loop, $120-3 \times 10-3 \mathrm{R}=0$ $3 \mathrm{R}=90$ $\mathrm{R}=30 \Omega$
