Physics, 31.03.2020 04:42 bferguson7986

2 resistors of resistance R1 and R2, with R2> R1, are connected to a voltage source with voltage V0. When the resistors are connected in series the current is Is. when the resistors are connected in parallel, the current Ip from the source is equal to 10 Is.

Let r be the ration R1/R2. find r.

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Answer from: enrique764058

The value of r is = 0.127

Explanation:

Generally for a series connection the equivalent resistance is mathematically evaluated as

R_e = R_1 + R_2

Where are R_1 and R_2 are given in the question

The current is mathematically represented as

$I_s = \frac{V_o}{R_e}$

$= \frac{V_o}{R_1 +R_2} ----(1)$

Generally for a parallel connection the equivalent resistance is mathematically evaluated as

$R_e = \frac{1}{R_1} + \frac{1}{R_2}$

$=\frac{R_1 R_2}{R_1 + R_2}$

The current is mathematically represented as

$I_p = \frac{V_o}{R_e}$

$I_p =\frac{V_o}{\frac{R_1 R_2}{R_1 +R_2} }$

$= \frac{V_o(R_1 +R_2)}{R_1 R_2} -----(2)$

Now we are told that I_p = 10I_s

=> $\frac{V_o(R-1 +R_2)}{R_1R_2} = 10 [\frac{V_o}{(R_1 +R_2)} ]$

=> $\frac{(R_1 + R_2)^2}{R_1 R_2} = 10$

=> $\frac{R_1^2 +R_1^2 + 2 R_1 R_2}{R_1 R_2 } = 10$

From the question $r = \frac{R_1}{R_2}$

Substituting this into the equation we have

$r + \frac{1}{r} =10-2$

Multiplying through by r

r^2 -8r +1 =0

Solving the equation using quadratic formula $\frac{-b \pm\sqrt{b^2 -4ac} }{2a}$

$r = \frac{8+ \sqrt{8^2 -4} }{2} \ or \ r= \frac{8 -\sqrt{8^2-4} }{2}$

Now recall the R_2 R_1 so r < 1

Hence r = 0.127

Answer from: niyawilliams23

The value for r = 0.12

Explanation:

In series combination, we have:

Req = R1 + R2

The current is:

$I_{s} =\frac{V_{0} }{R_{eq} } =\frac{V_{0} }{R_{1}+R_{2} }$ (eq. 1)

In parallel combination, we have:

$R_{eq}= \frac{R_{1}R_{2}}{R_{1}+R_{2}}$

The current is:

$I_{p} =\frac{V_{0}(R_{1}+R_{2})}{R_{1}R_{2}}$ (eq. 2)

If Ip = 10Is

$\frac{V_{0}(R_{1}+R_{2})}{R_{1}R_{2} } =10(\frac{V_{0}}{R_{1}+R_{2}} )\\\frac{(R_{1}+R_{2})^{2} }{R_{1}R_{2}} =10\\\frac{R_{1}^{2}+R_{2}^{2}+2R_{1}R_{2} }{R_{1}R_{2}} =10\\\frac{R_{1}}{R_{2}} +\frac{R_{2}}{R_{1}} +2=10\\$

if Y = r

$1+(\frac{1}{Y} )=8\\Y^{2}-8Y+1=0$

Solving:

$Y=\frac{8+-\sqrt{60} }{2} \\Y=\frac{R_{1}}{R_{2}} =0.12$

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