152685
Two electric bulbs $A$ and $b$ are rated as $60 \mathrm{~W}$ and $100 \mathrm{~W}$. They are connected in parallel to the same source. Then :
1 B draws more current than A
2 current drawn are in the ratio of their resistances
3 both draw the same current
4 A draws more current than B.
Explanation:
A Given, Power of electric bulb, $A=60 \mathrm{~W}$ Power of electric bulb, $B=100 \mathrm{~W}$ We know that, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Now, the resistance of $60 \mathrm{~W}$ bulb has more resistance as compared to a $100 \mathrm{~W}$ bulb as $\mathrm{P} \propto \frac{1}{\mathrm{R}}$. Therefore, when bulbs are connected in parallel to the same source, the bulb $\mathrm{B}(100 \mathrm{~W})$ draws more current than A.
Karnataka CET-2004
Current Electricity
152686
A fuse wire with radius $1 \mathrm{~mm}$ blows at 1.5 ampere. The radius of the fuse wire of the same material to blow at $3 \mathrm{~A}$ will be :
1 $3^{1 / 4} \mathrm{~mm}$
2 $4^{1 / 3} \mathrm{~mm}$
3 $3^{1 / 2} \mathrm{~mm}$
4 $2^{1 / 3} \mathrm{~mm}$
Explanation:
B Given, $\mathrm{r}_{1}=1 \mathrm{~mm}, \mathrm{I}_{1}=1.5 \mathrm{~A}, \mathrm{I}_{2}=3 \mathrm{~A}$ We know that, For fusing current, The relation of current with radius, $I^{2} \propto r^{3}$ $\frac{I_{1}^{2}}{r_{1}^{3}}=\frac{I_{2}^{2}}{r_{2}^{3}}$ $\left(\frac{I_{2}}{I_{1}}\right)^{2}=\left(\frac{r_{2}}{r_{1}}\right)^{3}$ $r_{2}=r_{1}\left(\frac{I_{2}}{I_{1}}\right)^{2 / 3}$ $r_{2}=1\left(\frac{3}{1.5}\right)^{2 / 3}$ $r_{2}=2^{2 / 3}=4^{1 / 3} \mathrm{~mm}$
Karnataka CET-2003
Current Electricity
152687
$n$ identical bulbs, each designated to draw a power $P$ from a certain voltage supply. The total power which they will draw is :
1 $P / n$
2 $\mathrm{P} / \mathrm{n}^{2}$
3 $\mathrm{nP}$
4 $\mathrm{P}$
Explanation:
A Resistance of each bulb $=\mathrm{R}$ Voltage of source $=\mathrm{V}$ We know that, Power of each bulb, $P=\frac{V^{2}}{R}$ $\text { or } \quad \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now, $\mathrm{n}$ bulbs are in series so total resistance, $\mathrm{R}_{\mathrm{t}}=\mathrm{nR}$ $\therefore \text { Current }(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{t}}}=\frac{\mathrm{V}}{\mathrm{nR}}$ Therefore, Total power drawn, $\mathrm{P}_{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{t}}$ $P_{t}=\frac{V^{2}}{n^{2} R^{2}}(n R)$ $P_{t}=\frac{V^{2}}{n R}=\frac{P}{n}$
Karnataka CET-2002
Current Electricity
152688
An electric bulb is designed to draw power $P_{0}$ at voltage $V_{0}$. If the voltage is $V$, it draws a power $P$. Then :
B We know that, Power in resistor, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ So, $\quad \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now since the bulb is operating initially at power $\mathrm{P}_{0}$ and voltage $\mathrm{V}_{0}$. So, $\quad \mathrm{P}_{0}=\frac{\mathrm{V}_{0}^{2}}{\mathrm{R}}$ Resistance of bulb $(\mathrm{R})=\frac{\mathrm{V}_{0}^{2}}{\mathrm{P}_{0}}$ Now the bulb is attached to voltage $\mathrm{V}$ and it consumes power P. $\text { So, } \mathrm{P} =\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{\mathrm{V}^{2} \mathrm{P}_{0}}{\mathrm{~V}_{0}^{2}}$ $\mathrm{P} =\left(\frac{\mathrm{V}}{\mathrm{V}_{0}}\right)^{2} \mathrm{P}_{0}$
Karnataka CET-2001
Current Electricity
152689
An electric bulb is rated at $220 \mathrm{~V}, 200 \mathrm{~W}$. Power consumed by it when operated at $110 \mathrm{~V}$ is
152685
Two electric bulbs $A$ and $b$ are rated as $60 \mathrm{~W}$ and $100 \mathrm{~W}$. They are connected in parallel to the same source. Then :
1 B draws more current than A
2 current drawn are in the ratio of their resistances
3 both draw the same current
4 A draws more current than B.
Explanation:
A Given, Power of electric bulb, $A=60 \mathrm{~W}$ Power of electric bulb, $B=100 \mathrm{~W}$ We know that, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Now, the resistance of $60 \mathrm{~W}$ bulb has more resistance as compared to a $100 \mathrm{~W}$ bulb as $\mathrm{P} \propto \frac{1}{\mathrm{R}}$. Therefore, when bulbs are connected in parallel to the same source, the bulb $\mathrm{B}(100 \mathrm{~W})$ draws more current than A.
Karnataka CET-2004
Current Electricity
152686
A fuse wire with radius $1 \mathrm{~mm}$ blows at 1.5 ampere. The radius of the fuse wire of the same material to blow at $3 \mathrm{~A}$ will be :
1 $3^{1 / 4} \mathrm{~mm}$
2 $4^{1 / 3} \mathrm{~mm}$
3 $3^{1 / 2} \mathrm{~mm}$
4 $2^{1 / 3} \mathrm{~mm}$
Explanation:
B Given, $\mathrm{r}_{1}=1 \mathrm{~mm}, \mathrm{I}_{1}=1.5 \mathrm{~A}, \mathrm{I}_{2}=3 \mathrm{~A}$ We know that, For fusing current, The relation of current with radius, $I^{2} \propto r^{3}$ $\frac{I_{1}^{2}}{r_{1}^{3}}=\frac{I_{2}^{2}}{r_{2}^{3}}$ $\left(\frac{I_{2}}{I_{1}}\right)^{2}=\left(\frac{r_{2}}{r_{1}}\right)^{3}$ $r_{2}=r_{1}\left(\frac{I_{2}}{I_{1}}\right)^{2 / 3}$ $r_{2}=1\left(\frac{3}{1.5}\right)^{2 / 3}$ $r_{2}=2^{2 / 3}=4^{1 / 3} \mathrm{~mm}$
Karnataka CET-2003
Current Electricity
152687
$n$ identical bulbs, each designated to draw a power $P$ from a certain voltage supply. The total power which they will draw is :
1 $P / n$
2 $\mathrm{P} / \mathrm{n}^{2}$
3 $\mathrm{nP}$
4 $\mathrm{P}$
Explanation:
A Resistance of each bulb $=\mathrm{R}$ Voltage of source $=\mathrm{V}$ We know that, Power of each bulb, $P=\frac{V^{2}}{R}$ $\text { or } \quad \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now, $\mathrm{n}$ bulbs are in series so total resistance, $\mathrm{R}_{\mathrm{t}}=\mathrm{nR}$ $\therefore \text { Current }(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{t}}}=\frac{\mathrm{V}}{\mathrm{nR}}$ Therefore, Total power drawn, $\mathrm{P}_{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{t}}$ $P_{t}=\frac{V^{2}}{n^{2} R^{2}}(n R)$ $P_{t}=\frac{V^{2}}{n R}=\frac{P}{n}$
Karnataka CET-2002
Current Electricity
152688
An electric bulb is designed to draw power $P_{0}$ at voltage $V_{0}$. If the voltage is $V$, it draws a power $P$. Then :
B We know that, Power in resistor, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ So, $\quad \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now since the bulb is operating initially at power $\mathrm{P}_{0}$ and voltage $\mathrm{V}_{0}$. So, $\quad \mathrm{P}_{0}=\frac{\mathrm{V}_{0}^{2}}{\mathrm{R}}$ Resistance of bulb $(\mathrm{R})=\frac{\mathrm{V}_{0}^{2}}{\mathrm{P}_{0}}$ Now the bulb is attached to voltage $\mathrm{V}$ and it consumes power P. $\text { So, } \mathrm{P} =\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{\mathrm{V}^{2} \mathrm{P}_{0}}{\mathrm{~V}_{0}^{2}}$ $\mathrm{P} =\left(\frac{\mathrm{V}}{\mathrm{V}_{0}}\right)^{2} \mathrm{P}_{0}$
Karnataka CET-2001
Current Electricity
152689
An electric bulb is rated at $220 \mathrm{~V}, 200 \mathrm{~W}$. Power consumed by it when operated at $110 \mathrm{~V}$ is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
152685
Two electric bulbs $A$ and $b$ are rated as $60 \mathrm{~W}$ and $100 \mathrm{~W}$. They are connected in parallel to the same source. Then :
1 B draws more current than A
2 current drawn are in the ratio of their resistances
3 both draw the same current
4 A draws more current than B.
Explanation:
A Given, Power of electric bulb, $A=60 \mathrm{~W}$ Power of electric bulb, $B=100 \mathrm{~W}$ We know that, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Now, the resistance of $60 \mathrm{~W}$ bulb has more resistance as compared to a $100 \mathrm{~W}$ bulb as $\mathrm{P} \propto \frac{1}{\mathrm{R}}$. Therefore, when bulbs are connected in parallel to the same source, the bulb $\mathrm{B}(100 \mathrm{~W})$ draws more current than A.
Karnataka CET-2004
Current Electricity
152686
A fuse wire with radius $1 \mathrm{~mm}$ blows at 1.5 ampere. The radius of the fuse wire of the same material to blow at $3 \mathrm{~A}$ will be :
1 $3^{1 / 4} \mathrm{~mm}$
2 $4^{1 / 3} \mathrm{~mm}$
3 $3^{1 / 2} \mathrm{~mm}$
4 $2^{1 / 3} \mathrm{~mm}$
Explanation:
B Given, $\mathrm{r}_{1}=1 \mathrm{~mm}, \mathrm{I}_{1}=1.5 \mathrm{~A}, \mathrm{I}_{2}=3 \mathrm{~A}$ We know that, For fusing current, The relation of current with radius, $I^{2} \propto r^{3}$ $\frac{I_{1}^{2}}{r_{1}^{3}}=\frac{I_{2}^{2}}{r_{2}^{3}}$ $\left(\frac{I_{2}}{I_{1}}\right)^{2}=\left(\frac{r_{2}}{r_{1}}\right)^{3}$ $r_{2}=r_{1}\left(\frac{I_{2}}{I_{1}}\right)^{2 / 3}$ $r_{2}=1\left(\frac{3}{1.5}\right)^{2 / 3}$ $r_{2}=2^{2 / 3}=4^{1 / 3} \mathrm{~mm}$
Karnataka CET-2003
Current Electricity
152687
$n$ identical bulbs, each designated to draw a power $P$ from a certain voltage supply. The total power which they will draw is :
1 $P / n$
2 $\mathrm{P} / \mathrm{n}^{2}$
3 $\mathrm{nP}$
4 $\mathrm{P}$
Explanation:
A Resistance of each bulb $=\mathrm{R}$ Voltage of source $=\mathrm{V}$ We know that, Power of each bulb, $P=\frac{V^{2}}{R}$ $\text { or } \quad \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now, $\mathrm{n}$ bulbs are in series so total resistance, $\mathrm{R}_{\mathrm{t}}=\mathrm{nR}$ $\therefore \text { Current }(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{t}}}=\frac{\mathrm{V}}{\mathrm{nR}}$ Therefore, Total power drawn, $\mathrm{P}_{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{t}}$ $P_{t}=\frac{V^{2}}{n^{2} R^{2}}(n R)$ $P_{t}=\frac{V^{2}}{n R}=\frac{P}{n}$
Karnataka CET-2002
Current Electricity
152688
An electric bulb is designed to draw power $P_{0}$ at voltage $V_{0}$. If the voltage is $V$, it draws a power $P$. Then :
B We know that, Power in resistor, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ So, $\quad \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now since the bulb is operating initially at power $\mathrm{P}_{0}$ and voltage $\mathrm{V}_{0}$. So, $\quad \mathrm{P}_{0}=\frac{\mathrm{V}_{0}^{2}}{\mathrm{R}}$ Resistance of bulb $(\mathrm{R})=\frac{\mathrm{V}_{0}^{2}}{\mathrm{P}_{0}}$ Now the bulb is attached to voltage $\mathrm{V}$ and it consumes power P. $\text { So, } \mathrm{P} =\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{\mathrm{V}^{2} \mathrm{P}_{0}}{\mathrm{~V}_{0}^{2}}$ $\mathrm{P} =\left(\frac{\mathrm{V}}{\mathrm{V}_{0}}\right)^{2} \mathrm{P}_{0}$
Karnataka CET-2001
Current Electricity
152689
An electric bulb is rated at $220 \mathrm{~V}, 200 \mathrm{~W}$. Power consumed by it when operated at $110 \mathrm{~V}$ is
152685
Two electric bulbs $A$ and $b$ are rated as $60 \mathrm{~W}$ and $100 \mathrm{~W}$. They are connected in parallel to the same source. Then :
1 B draws more current than A
2 current drawn are in the ratio of their resistances
3 both draw the same current
4 A draws more current than B.
Explanation:
A Given, Power of electric bulb, $A=60 \mathrm{~W}$ Power of electric bulb, $B=100 \mathrm{~W}$ We know that, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Now, the resistance of $60 \mathrm{~W}$ bulb has more resistance as compared to a $100 \mathrm{~W}$ bulb as $\mathrm{P} \propto \frac{1}{\mathrm{R}}$. Therefore, when bulbs are connected in parallel to the same source, the bulb $\mathrm{B}(100 \mathrm{~W})$ draws more current than A.
Karnataka CET-2004
Current Electricity
152686
A fuse wire with radius $1 \mathrm{~mm}$ blows at 1.5 ampere. The radius of the fuse wire of the same material to blow at $3 \mathrm{~A}$ will be :
1 $3^{1 / 4} \mathrm{~mm}$
2 $4^{1 / 3} \mathrm{~mm}$
3 $3^{1 / 2} \mathrm{~mm}$
4 $2^{1 / 3} \mathrm{~mm}$
Explanation:
B Given, $\mathrm{r}_{1}=1 \mathrm{~mm}, \mathrm{I}_{1}=1.5 \mathrm{~A}, \mathrm{I}_{2}=3 \mathrm{~A}$ We know that, For fusing current, The relation of current with radius, $I^{2} \propto r^{3}$ $\frac{I_{1}^{2}}{r_{1}^{3}}=\frac{I_{2}^{2}}{r_{2}^{3}}$ $\left(\frac{I_{2}}{I_{1}}\right)^{2}=\left(\frac{r_{2}}{r_{1}}\right)^{3}$ $r_{2}=r_{1}\left(\frac{I_{2}}{I_{1}}\right)^{2 / 3}$ $r_{2}=1\left(\frac{3}{1.5}\right)^{2 / 3}$ $r_{2}=2^{2 / 3}=4^{1 / 3} \mathrm{~mm}$
Karnataka CET-2003
Current Electricity
152687
$n$ identical bulbs, each designated to draw a power $P$ from a certain voltage supply. The total power which they will draw is :
1 $P / n$
2 $\mathrm{P} / \mathrm{n}^{2}$
3 $\mathrm{nP}$
4 $\mathrm{P}$
Explanation:
A Resistance of each bulb $=\mathrm{R}$ Voltage of source $=\mathrm{V}$ We know that, Power of each bulb, $P=\frac{V^{2}}{R}$ $\text { or } \quad \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now, $\mathrm{n}$ bulbs are in series so total resistance, $\mathrm{R}_{\mathrm{t}}=\mathrm{nR}$ $\therefore \text { Current }(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{t}}}=\frac{\mathrm{V}}{\mathrm{nR}}$ Therefore, Total power drawn, $\mathrm{P}_{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{t}}$ $P_{t}=\frac{V^{2}}{n^{2} R^{2}}(n R)$ $P_{t}=\frac{V^{2}}{n R}=\frac{P}{n}$
Karnataka CET-2002
Current Electricity
152688
An electric bulb is designed to draw power $P_{0}$ at voltage $V_{0}$. If the voltage is $V$, it draws a power $P$. Then :
B We know that, Power in resistor, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ So, $\quad \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now since the bulb is operating initially at power $\mathrm{P}_{0}$ and voltage $\mathrm{V}_{0}$. So, $\quad \mathrm{P}_{0}=\frac{\mathrm{V}_{0}^{2}}{\mathrm{R}}$ Resistance of bulb $(\mathrm{R})=\frac{\mathrm{V}_{0}^{2}}{\mathrm{P}_{0}}$ Now the bulb is attached to voltage $\mathrm{V}$ and it consumes power P. $\text { So, } \mathrm{P} =\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{\mathrm{V}^{2} \mathrm{P}_{0}}{\mathrm{~V}_{0}^{2}}$ $\mathrm{P} =\left(\frac{\mathrm{V}}{\mathrm{V}_{0}}\right)^{2} \mathrm{P}_{0}$
Karnataka CET-2001
Current Electricity
152689
An electric bulb is rated at $220 \mathrm{~V}, 200 \mathrm{~W}$. Power consumed by it when operated at $110 \mathrm{~V}$ is
152685
Two electric bulbs $A$ and $b$ are rated as $60 \mathrm{~W}$ and $100 \mathrm{~W}$. They are connected in parallel to the same source. Then :
1 B draws more current than A
2 current drawn are in the ratio of their resistances
3 both draw the same current
4 A draws more current than B.
Explanation:
A Given, Power of electric bulb, $A=60 \mathrm{~W}$ Power of electric bulb, $B=100 \mathrm{~W}$ We know that, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Now, the resistance of $60 \mathrm{~W}$ bulb has more resistance as compared to a $100 \mathrm{~W}$ bulb as $\mathrm{P} \propto \frac{1}{\mathrm{R}}$. Therefore, when bulbs are connected in parallel to the same source, the bulb $\mathrm{B}(100 \mathrm{~W})$ draws more current than A.
Karnataka CET-2004
Current Electricity
152686
A fuse wire with radius $1 \mathrm{~mm}$ blows at 1.5 ampere. The radius of the fuse wire of the same material to blow at $3 \mathrm{~A}$ will be :
1 $3^{1 / 4} \mathrm{~mm}$
2 $4^{1 / 3} \mathrm{~mm}$
3 $3^{1 / 2} \mathrm{~mm}$
4 $2^{1 / 3} \mathrm{~mm}$
Explanation:
B Given, $\mathrm{r}_{1}=1 \mathrm{~mm}, \mathrm{I}_{1}=1.5 \mathrm{~A}, \mathrm{I}_{2}=3 \mathrm{~A}$ We know that, For fusing current, The relation of current with radius, $I^{2} \propto r^{3}$ $\frac{I_{1}^{2}}{r_{1}^{3}}=\frac{I_{2}^{2}}{r_{2}^{3}}$ $\left(\frac{I_{2}}{I_{1}}\right)^{2}=\left(\frac{r_{2}}{r_{1}}\right)^{3}$ $r_{2}=r_{1}\left(\frac{I_{2}}{I_{1}}\right)^{2 / 3}$ $r_{2}=1\left(\frac{3}{1.5}\right)^{2 / 3}$ $r_{2}=2^{2 / 3}=4^{1 / 3} \mathrm{~mm}$
Karnataka CET-2003
Current Electricity
152687
$n$ identical bulbs, each designated to draw a power $P$ from a certain voltage supply. The total power which they will draw is :
1 $P / n$
2 $\mathrm{P} / \mathrm{n}^{2}$
3 $\mathrm{nP}$
4 $\mathrm{P}$
Explanation:
A Resistance of each bulb $=\mathrm{R}$ Voltage of source $=\mathrm{V}$ We know that, Power of each bulb, $P=\frac{V^{2}}{R}$ $\text { or } \quad \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now, $\mathrm{n}$ bulbs are in series so total resistance, $\mathrm{R}_{\mathrm{t}}=\mathrm{nR}$ $\therefore \text { Current }(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{t}}}=\frac{\mathrm{V}}{\mathrm{nR}}$ Therefore, Total power drawn, $\mathrm{P}_{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{t}}$ $P_{t}=\frac{V^{2}}{n^{2} R^{2}}(n R)$ $P_{t}=\frac{V^{2}}{n R}=\frac{P}{n}$
Karnataka CET-2002
Current Electricity
152688
An electric bulb is designed to draw power $P_{0}$ at voltage $V_{0}$. If the voltage is $V$, it draws a power $P$. Then :
B We know that, Power in resistor, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ So, $\quad \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now since the bulb is operating initially at power $\mathrm{P}_{0}$ and voltage $\mathrm{V}_{0}$. So, $\quad \mathrm{P}_{0}=\frac{\mathrm{V}_{0}^{2}}{\mathrm{R}}$ Resistance of bulb $(\mathrm{R})=\frac{\mathrm{V}_{0}^{2}}{\mathrm{P}_{0}}$ Now the bulb is attached to voltage $\mathrm{V}$ and it consumes power P. $\text { So, } \mathrm{P} =\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{\mathrm{V}^{2} \mathrm{P}_{0}}{\mathrm{~V}_{0}^{2}}$ $\mathrm{P} =\left(\frac{\mathrm{V}}{\mathrm{V}_{0}}\right)^{2} \mathrm{P}_{0}$
Karnataka CET-2001
Current Electricity
152689
An electric bulb is rated at $220 \mathrm{~V}, 200 \mathrm{~W}$. Power consumed by it when operated at $110 \mathrm{~V}$ is