Question

please answer the equivalent resistance of this​

Answer 1

Answer:

total resistance of wire A

$\</strong><strong>g</strong><strong>r</strong><strong>e</strong><strong>e</strong><strong>n</strong><strong>{</strong><strong>(</strong><strong>1</strong><strong>0</strong><strong>+</strong><strong>1</strong><strong>0</strong><strong>+</strong><strong>1</strong><strong>0</strong><strong>)</strong><strong>ohm</strong><strong>=</strong><strong>3</strong><strong>0</strong><strong>o</strong><strong>h</strong><strong>m</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>(</strong><strong>1</strong><strong>)</strong><strong>}$

total resistance of wire B

$\</strong><strong>g</strong><strong>r</strong><strong>e</strong><strong>e</strong><strong>n</strong><strong>{</strong><strong>(</strong><strong>1</strong><strong>0</strong><strong>+</strong><strong>1</strong><strong>0</strong><strong>+</strong><strong>1</strong><strong>0</strong><strong>)</strong><strong>ohm</strong><strong>=</strong><strong>3</strong><strong>0</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>(</strong><strong>2</strong><strong>)</strong><strong>}$

equivalent resistance of the wires CE and Df connected parallel

$\</strong><strong>g</strong><strong>r</strong><strong>e</strong><strong>e</strong><strong>n</strong><strong>{</strong><strong>R'</strong><strong>=</strong><strong>1</strong><strong>/</strong><strong>1</strong><strong>0</strong><strong>+</strong><strong>1</strong><strong>/</strong><strong>1</strong><strong>0</strong><strong>}$

$\</strong><strong>g</strong><strong>r</strong><strong>e</strong><strong>e</strong><strong>n</strong><strong>{</strong><strong>R'</strong><strong>=</strong><strong>2</strong><strong>/</strong><strong>1</strong><strong>0</strong><strong>=</strong><strong>1</strong><strong>/</strong><strong>5</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>(</strong><strong>3</strong><strong>)</strong><strong>}$

adding 1,2 and 3 we get the equilvalent resistance

$\large{</strong><strong>=</strong><strong>(</strong><strong>3</strong><strong>0</strong><strong>+</strong><strong>3</strong><strong>0</strong><strong>+</strong><strong>1</strong><strong>/</strong><strong>5</strong><strong>)</strong><strong>ohm</strong><strong>}$

$\large{</strong><strong>=</strong><strong>(</strong><strong>1</strong><strong>5</strong><strong>0</strong><strong>+</strong><strong>1</strong><strong>5</strong><strong>0</strong><strong>+</strong><strong>1</strong><strong>/</strong><strong>5</strong><strong>)</strong><strong>ohm</strong><strong>}$

$\large{</strong><strong>=</strong><strong>3</strong><strong>0</strong><strong>1</strong><strong>/</strong><strong>5</strong><strong>o</strong><strong>h</strong><strong>m</strong><strong>}$

$\large{</strong><strong>=</strong><strong>6</strong><strong>0</strong><strong>.</strong><strong>2</strong><strong>o</strong><strong>h</strong><strong>m</strong><strong>}$