152767
Two electric resistors have equal values of resistance $R$. Each can be operated with a power of 320 watts (W) at 220 volts. If the two resistors are connected in series to a 110 volts electric supply, then the power generated in each resistor is
1 90 watts
2 81 watts
3 60 watts
4 20 watts
Explanation:
D Given, Two electric resistor have equal values of resistance $R$ Power $(\mathrm{P})=320 \mathrm{~W}$ $\mathrm{V}_{1}=220 \text { volts }$ $\mathrm{V}_{2}=110 \text { volts }$ We know that, Power, $\mathrm{P}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{320}$ Resistance, $\mathrm{R}=151.25 \Omega$ If the two resistors are connected in series. Then, Total resistance $\left(\mathrm{R}^{\prime}\right)=\mathrm{R}+\mathrm{R}$ $\mathrm{R}^{\prime}=2 \mathrm{R}$ $\mathrm{R}^{\prime}=2 \times 151.25$ $\mathrm{R}^{\prime}=302.50 \Omega$ $\therefore \quad \mathrm{P}^{\prime} =\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}^{\prime}}=\frac{(110)^{2}}{302.50}$ $\mathrm{P}^{\prime} =40 \mathrm{~W}$ Then the power generated in each resistor is $\mathrm{P}^{\prime \prime} =\frac{40}{2}$ $\mathrm{P}^{\prime \prime} =20 \mathrm{~W}$
TS EAMCET (Engg.)-2016
Current Electricity
152768
The resistors of $6 \Omega$ and $9 \Omega$ are connected in series to a $120 \mathrm{~V}$ source. The power consumed by $6 \Omega$ resistor is :
1 $384 \mathrm{~W}$
2 $606 \mathrm{~W}$
3 $1500 \mathrm{~W}$
4 $1800 \mathrm{~W}$
Explanation:
A Given, $\mathrm{R}_{1}=6 \Omega$ $\mathrm{R}_{2}=9 \Omega$ $\mathrm{V}=120 \text { volts }$ $\therefore \quad \mathrm{R}^{\prime}=\mathrm{R}_{1}+\mathrm{R}_{2}=6+9$ $\mathrm{R}^{\prime}=15 \Omega$ According to Ohm's law, $\frac{\mathrm{V}}{\mathrm{I}}=\mathrm{R}$ $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{120}{15}$ $\mathrm{I}=8 \mathrm{~A}$ Power consumed in $6 \Omega$ resistance, $\text { Power } (\mathrm{P}) =\mathrm{I}^{2} \mathrm{R}$ $\mathrm{P} =(8)^{2} \times 6 \quad[\therefore \mathrm{R}=6 \Omega]$ $\mathrm{P} =64 \times 6$ $\mathrm{P} =384 \mathrm{~W}$
MP PMT-2013
Current Electricity
152769
A heater is marked 500 watt, 200 volts. The cost of using the heater for four hours at 15 paisa per unit is :
1 90 paisa
2 60 paisa
3 30 paisa
4 15 paisa
Explanation:
C Given, $\mathrm{P}=500 \text { watt }$ $\mathrm{V}=200 \text { volts }$ Time $(\mathrm{t})=4$ hours We know that, $\text { Energy } =\text { power } \times \text { time }$ $=500 \times 4$ $=2000$ $=2 \mathrm{kWh}$ As $1 \mathrm{kWh}$ is considered as 1 unit So, cost $=2$ unit $\times 15$ paisa/unit $=30 \text { paisa }$
MP PET -2013
Current Electricity
152770
Two bulbs consume the same power when operated at $200 \mathrm{~V}$ and $300 \mathrm{~V}$ respectively. When these bulbs are connected in series across a D.C. source of $500 \mathrm{~V}$ then the ratio of potential difference across them is
1 $\frac{2}{3}$
2 $\frac{4}{9}$
3 $\frac{6}{27}$
4 $\frac{8}{24}$
Explanation:
B According to question, $\mathrm{P}_{1}=\mathrm{P}_{2}$ Power, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\therefore \quad \frac{\mathrm{V}_{1}^{2}}{\mathrm{R}_{1}}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}_{2}} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}_{2}\right]$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{~V}_{2}^{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{(200)^{2}}{(300)^{2}}=\frac{4}{9}$ Now, bulbs are connected in series then current will be same in both bulbs. Or $\quad \mathrm{I}_{1}=\mathrm{I}_{2}$ $\frac{\mathrm{V}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{R}_{2}}$ The ratio of potential difference $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{4}{9}$
152767
Two electric resistors have equal values of resistance $R$. Each can be operated with a power of 320 watts (W) at 220 volts. If the two resistors are connected in series to a 110 volts electric supply, then the power generated in each resistor is
1 90 watts
2 81 watts
3 60 watts
4 20 watts
Explanation:
D Given, Two electric resistor have equal values of resistance $R$ Power $(\mathrm{P})=320 \mathrm{~W}$ $\mathrm{V}_{1}=220 \text { volts }$ $\mathrm{V}_{2}=110 \text { volts }$ We know that, Power, $\mathrm{P}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{320}$ Resistance, $\mathrm{R}=151.25 \Omega$ If the two resistors are connected in series. Then, Total resistance $\left(\mathrm{R}^{\prime}\right)=\mathrm{R}+\mathrm{R}$ $\mathrm{R}^{\prime}=2 \mathrm{R}$ $\mathrm{R}^{\prime}=2 \times 151.25$ $\mathrm{R}^{\prime}=302.50 \Omega$ $\therefore \quad \mathrm{P}^{\prime} =\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}^{\prime}}=\frac{(110)^{2}}{302.50}$ $\mathrm{P}^{\prime} =40 \mathrm{~W}$ Then the power generated in each resistor is $\mathrm{P}^{\prime \prime} =\frac{40}{2}$ $\mathrm{P}^{\prime \prime} =20 \mathrm{~W}$
TS EAMCET (Engg.)-2016
Current Electricity
152768
The resistors of $6 \Omega$ and $9 \Omega$ are connected in series to a $120 \mathrm{~V}$ source. The power consumed by $6 \Omega$ resistor is :
1 $384 \mathrm{~W}$
2 $606 \mathrm{~W}$
3 $1500 \mathrm{~W}$
4 $1800 \mathrm{~W}$
Explanation:
A Given, $\mathrm{R}_{1}=6 \Omega$ $\mathrm{R}_{2}=9 \Omega$ $\mathrm{V}=120 \text { volts }$ $\therefore \quad \mathrm{R}^{\prime}=\mathrm{R}_{1}+\mathrm{R}_{2}=6+9$ $\mathrm{R}^{\prime}=15 \Omega$ According to Ohm's law, $\frac{\mathrm{V}}{\mathrm{I}}=\mathrm{R}$ $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{120}{15}$ $\mathrm{I}=8 \mathrm{~A}$ Power consumed in $6 \Omega$ resistance, $\text { Power } (\mathrm{P}) =\mathrm{I}^{2} \mathrm{R}$ $\mathrm{P} =(8)^{2} \times 6 \quad[\therefore \mathrm{R}=6 \Omega]$ $\mathrm{P} =64 \times 6$ $\mathrm{P} =384 \mathrm{~W}$
MP PMT-2013
Current Electricity
152769
A heater is marked 500 watt, 200 volts. The cost of using the heater for four hours at 15 paisa per unit is :
1 90 paisa
2 60 paisa
3 30 paisa
4 15 paisa
Explanation:
C Given, $\mathrm{P}=500 \text { watt }$ $\mathrm{V}=200 \text { volts }$ Time $(\mathrm{t})=4$ hours We know that, $\text { Energy } =\text { power } \times \text { time }$ $=500 \times 4$ $=2000$ $=2 \mathrm{kWh}$ As $1 \mathrm{kWh}$ is considered as 1 unit So, cost $=2$ unit $\times 15$ paisa/unit $=30 \text { paisa }$
MP PET -2013
Current Electricity
152770
Two bulbs consume the same power when operated at $200 \mathrm{~V}$ and $300 \mathrm{~V}$ respectively. When these bulbs are connected in series across a D.C. source of $500 \mathrm{~V}$ then the ratio of potential difference across them is
1 $\frac{2}{3}$
2 $\frac{4}{9}$
3 $\frac{6}{27}$
4 $\frac{8}{24}$
Explanation:
B According to question, $\mathrm{P}_{1}=\mathrm{P}_{2}$ Power, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\therefore \quad \frac{\mathrm{V}_{1}^{2}}{\mathrm{R}_{1}}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}_{2}} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}_{2}\right]$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{~V}_{2}^{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{(200)^{2}}{(300)^{2}}=\frac{4}{9}$ Now, bulbs are connected in series then current will be same in both bulbs. Or $\quad \mathrm{I}_{1}=\mathrm{I}_{2}$ $\frac{\mathrm{V}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{R}_{2}}$ The ratio of potential difference $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{4}{9}$
152767
Two electric resistors have equal values of resistance $R$. Each can be operated with a power of 320 watts (W) at 220 volts. If the two resistors are connected in series to a 110 volts electric supply, then the power generated in each resistor is
1 90 watts
2 81 watts
3 60 watts
4 20 watts
Explanation:
D Given, Two electric resistor have equal values of resistance $R$ Power $(\mathrm{P})=320 \mathrm{~W}$ $\mathrm{V}_{1}=220 \text { volts }$ $\mathrm{V}_{2}=110 \text { volts }$ We know that, Power, $\mathrm{P}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{320}$ Resistance, $\mathrm{R}=151.25 \Omega$ If the two resistors are connected in series. Then, Total resistance $\left(\mathrm{R}^{\prime}\right)=\mathrm{R}+\mathrm{R}$ $\mathrm{R}^{\prime}=2 \mathrm{R}$ $\mathrm{R}^{\prime}=2 \times 151.25$ $\mathrm{R}^{\prime}=302.50 \Omega$ $\therefore \quad \mathrm{P}^{\prime} =\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}^{\prime}}=\frac{(110)^{2}}{302.50}$ $\mathrm{P}^{\prime} =40 \mathrm{~W}$ Then the power generated in each resistor is $\mathrm{P}^{\prime \prime} =\frac{40}{2}$ $\mathrm{P}^{\prime \prime} =20 \mathrm{~W}$
TS EAMCET (Engg.)-2016
Current Electricity
152768
The resistors of $6 \Omega$ and $9 \Omega$ are connected in series to a $120 \mathrm{~V}$ source. The power consumed by $6 \Omega$ resistor is :
1 $384 \mathrm{~W}$
2 $606 \mathrm{~W}$
3 $1500 \mathrm{~W}$
4 $1800 \mathrm{~W}$
Explanation:
A Given, $\mathrm{R}_{1}=6 \Omega$ $\mathrm{R}_{2}=9 \Omega$ $\mathrm{V}=120 \text { volts }$ $\therefore \quad \mathrm{R}^{\prime}=\mathrm{R}_{1}+\mathrm{R}_{2}=6+9$ $\mathrm{R}^{\prime}=15 \Omega$ According to Ohm's law, $\frac{\mathrm{V}}{\mathrm{I}}=\mathrm{R}$ $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{120}{15}$ $\mathrm{I}=8 \mathrm{~A}$ Power consumed in $6 \Omega$ resistance, $\text { Power } (\mathrm{P}) =\mathrm{I}^{2} \mathrm{R}$ $\mathrm{P} =(8)^{2} \times 6 \quad[\therefore \mathrm{R}=6 \Omega]$ $\mathrm{P} =64 \times 6$ $\mathrm{P} =384 \mathrm{~W}$
MP PMT-2013
Current Electricity
152769
A heater is marked 500 watt, 200 volts. The cost of using the heater for four hours at 15 paisa per unit is :
1 90 paisa
2 60 paisa
3 30 paisa
4 15 paisa
Explanation:
C Given, $\mathrm{P}=500 \text { watt }$ $\mathrm{V}=200 \text { volts }$ Time $(\mathrm{t})=4$ hours We know that, $\text { Energy } =\text { power } \times \text { time }$ $=500 \times 4$ $=2000$ $=2 \mathrm{kWh}$ As $1 \mathrm{kWh}$ is considered as 1 unit So, cost $=2$ unit $\times 15$ paisa/unit $=30 \text { paisa }$
MP PET -2013
Current Electricity
152770
Two bulbs consume the same power when operated at $200 \mathrm{~V}$ and $300 \mathrm{~V}$ respectively. When these bulbs are connected in series across a D.C. source of $500 \mathrm{~V}$ then the ratio of potential difference across them is
1 $\frac{2}{3}$
2 $\frac{4}{9}$
3 $\frac{6}{27}$
4 $\frac{8}{24}$
Explanation:
B According to question, $\mathrm{P}_{1}=\mathrm{P}_{2}$ Power, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\therefore \quad \frac{\mathrm{V}_{1}^{2}}{\mathrm{R}_{1}}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}_{2}} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}_{2}\right]$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{~V}_{2}^{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{(200)^{2}}{(300)^{2}}=\frac{4}{9}$ Now, bulbs are connected in series then current will be same in both bulbs. Or $\quad \mathrm{I}_{1}=\mathrm{I}_{2}$ $\frac{\mathrm{V}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{R}_{2}}$ The ratio of potential difference $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{4}{9}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152767
Two electric resistors have equal values of resistance $R$. Each can be operated with a power of 320 watts (W) at 220 volts. If the two resistors are connected in series to a 110 volts electric supply, then the power generated in each resistor is
1 90 watts
2 81 watts
3 60 watts
4 20 watts
Explanation:
D Given, Two electric resistor have equal values of resistance $R$ Power $(\mathrm{P})=320 \mathrm{~W}$ $\mathrm{V}_{1}=220 \text { volts }$ $\mathrm{V}_{2}=110 \text { volts }$ We know that, Power, $\mathrm{P}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{320}$ Resistance, $\mathrm{R}=151.25 \Omega$ If the two resistors are connected in series. Then, Total resistance $\left(\mathrm{R}^{\prime}\right)=\mathrm{R}+\mathrm{R}$ $\mathrm{R}^{\prime}=2 \mathrm{R}$ $\mathrm{R}^{\prime}=2 \times 151.25$ $\mathrm{R}^{\prime}=302.50 \Omega$ $\therefore \quad \mathrm{P}^{\prime} =\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}^{\prime}}=\frac{(110)^{2}}{302.50}$ $\mathrm{P}^{\prime} =40 \mathrm{~W}$ Then the power generated in each resistor is $\mathrm{P}^{\prime \prime} =\frac{40}{2}$ $\mathrm{P}^{\prime \prime} =20 \mathrm{~W}$
TS EAMCET (Engg.)-2016
Current Electricity
152768
The resistors of $6 \Omega$ and $9 \Omega$ are connected in series to a $120 \mathrm{~V}$ source. The power consumed by $6 \Omega$ resistor is :
1 $384 \mathrm{~W}$
2 $606 \mathrm{~W}$
3 $1500 \mathrm{~W}$
4 $1800 \mathrm{~W}$
Explanation:
A Given, $\mathrm{R}_{1}=6 \Omega$ $\mathrm{R}_{2}=9 \Omega$ $\mathrm{V}=120 \text { volts }$ $\therefore \quad \mathrm{R}^{\prime}=\mathrm{R}_{1}+\mathrm{R}_{2}=6+9$ $\mathrm{R}^{\prime}=15 \Omega$ According to Ohm's law, $\frac{\mathrm{V}}{\mathrm{I}}=\mathrm{R}$ $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{120}{15}$ $\mathrm{I}=8 \mathrm{~A}$ Power consumed in $6 \Omega$ resistance, $\text { Power } (\mathrm{P}) =\mathrm{I}^{2} \mathrm{R}$ $\mathrm{P} =(8)^{2} \times 6 \quad[\therefore \mathrm{R}=6 \Omega]$ $\mathrm{P} =64 \times 6$ $\mathrm{P} =384 \mathrm{~W}$
MP PMT-2013
Current Electricity
152769
A heater is marked 500 watt, 200 volts. The cost of using the heater for four hours at 15 paisa per unit is :
1 90 paisa
2 60 paisa
3 30 paisa
4 15 paisa
Explanation:
C Given, $\mathrm{P}=500 \text { watt }$ $\mathrm{V}=200 \text { volts }$ Time $(\mathrm{t})=4$ hours We know that, $\text { Energy } =\text { power } \times \text { time }$ $=500 \times 4$ $=2000$ $=2 \mathrm{kWh}$ As $1 \mathrm{kWh}$ is considered as 1 unit So, cost $=2$ unit $\times 15$ paisa/unit $=30 \text { paisa }$
MP PET -2013
Current Electricity
152770
Two bulbs consume the same power when operated at $200 \mathrm{~V}$ and $300 \mathrm{~V}$ respectively. When these bulbs are connected in series across a D.C. source of $500 \mathrm{~V}$ then the ratio of potential difference across them is
1 $\frac{2}{3}$
2 $\frac{4}{9}$
3 $\frac{6}{27}$
4 $\frac{8}{24}$
Explanation:
B According to question, $\mathrm{P}_{1}=\mathrm{P}_{2}$ Power, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\therefore \quad \frac{\mathrm{V}_{1}^{2}}{\mathrm{R}_{1}}=\frac{\mathrm{V}_{2}^{2}}{\mathrm{R}_{2}} \quad\left[\because \mathrm{P}_{1}=\mathrm{P}_{2}\right]$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{~V}_{2}^{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{(200)^{2}}{(300)^{2}}=\frac{4}{9}$ Now, bulbs are connected in series then current will be same in both bulbs. Or $\quad \mathrm{I}_{1}=\mathrm{I}_{2}$ $\frac{\mathrm{V}_{1}}{\mathrm{R}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{R}_{2}}$ The ratio of potential difference $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{4}{9}$
